Energy and WorkEnergy and Work

Chapter 7Chapter 7

Introduction to EnergyIntroduction to Energy

The concept of energy is one of the The concept of energy is one of the most important topics in sciencemost important topics in science

Every physical process that occurs in Every physical process that occurs in the Universe involves energy and the Universe involves energy and energy transfers or transformationsenergy transfers or transformations

Work and EnergyWork and Energy

EnergyEnergy scalar quantity associated with a state (or scalar quantity associated with a state (or condition) of one or more objects condition) of one or more objects

Work and energy are scalars measured in NWork and energy are scalars measured in Nmiddotmiddotm or m or

Joules J 1J=kg∙mJoules J 1J=kg∙m22ss22

Energy can exist in many forms - mechanical Energy can exist in many forms - mechanical

electrical nuclear thermal chemicalhellipelectrical nuclear thermal chemicalhellip

Energy Approach to ProblemsEnergy Approach to Problems

The energy approach to describing The energy approach to describing motion is particularly useful when the motion is particularly useful when the force is not constantforce is not constant

An approach will involve An approach will involve Conservation Conservation of Energyof Energybull This could be extended to biological This could be extended to biological

organisms technological systems and organisms technological systems and engineering situationsengineering situations

Work and EnergyWork and Energy

Energy is a conserved quantity - the total Energy is a conserved quantity - the total amount of energy in the universe is constantamount of energy in the universe is constant

Energy can be converted from one type to Energy can be converted from one type to

another but never destroyedanother but never destroyed

Work and energy concepts can simplify Work and energy concepts can simplify

solutions of mechanical problems - they can solutions of mechanical problems - they can

be used in an alternative analysisbe used in an alternative analysis

SystemsSystems

A A systemsystem is a small portion of the is a small portion of the UniverseUniversebull We will ignore the details of the rest of the We will ignore the details of the rest of the

UniverseUniverse

A critical skill is to identify the systemA critical skill is to identify the system

Valid SystemValid System

A valid system mayA valid system maybull be a single object or particlebe a single object or particlebull be a collection of objects or particlesbe a collection of objects or particlesbull be a region of spacebe a region of spacebull vary in size and shapevary in size and shape

EnvironmentEnvironment

There is a There is a system boundarysystem boundary around the around the systemsystembull The boundary is an imaginary surfaceThe boundary is an imaginary surfacebull It does not necessarily correspond to a physical It does not necessarily correspond to a physical

boundaryboundary

The boundary divides the system from the The boundary divides the system from the environmentenvironmentbull The environment is the rest of the UniverseThe environment is the rest of the Universe

WorkWork

The workThe work WW done on a system by an done on a system by an agent exerting a constant force on the agent exerting a constant force on the system is the product of the magnitudesystem is the product of the magnitude FF of the of the forceforce the magnitude the magnitude ΔΔrr of the of the displacementdisplacement of the point of application of of the point of application of the force andthe force and cos cos θθ where where θθ is the angle is the angle between the force and the displacement between the force and the displacement vectorsvectors

WorkWork

W = F W = F ΔΔr cos r cos θθbull The displacement is that of the point of The displacement is that of the point of

application of the forceapplication of the forcebull A force does no work on the object if the A force does no work on the object if the

object does not move through a displacementobject does not move through a displacementbull The work done by a force on a moving object The work done by a force on a moving object

is zero when the force applied is is zero when the force applied is perpendicular to the displacement of its point perpendicular to the displacement of its point of applicationof application

Work ExampleWork Example

The normal forceThe normal force nn and and the gravitational forcethe gravitational force mgmg do no work on the objectdo no work on the object

cos cos θθ = cos 90deg= 0= cos 90deg= 0

The forceThe force FF does do work does do work on the objecton the object

More About WorkMore About Work The system and the environment must be The system and the environment must be

determined when dealing with workdetermined when dealing with work The environment does work on the systemThe environment does work on the system

(Work (Work byby the environment the environment onon the system) the system) The sign of the work depends on the direction ofThe sign of the work depends on the direction of

FF relative torelative to ΔΔrr Work is positive when projection ofWork is positive when projection of FF ontoonto ΔΔrr

is in the same direction as the displacementis in the same direction as the displacement Work is negative when the projection is in the Work is negative when the projection is in the

opposite directionopposite direction

Work Is An Energy TransferWork Is An Energy Transfer

bull This is important for a system approach This is important for a system approach to solving a problemto solving a problem

bull If the work is done on a system and it is If the work is done on a system and it is positive energy is transferred to the positive energy is transferred to the systemsystem

bull If the work done on the system is If the work done on the system is negative energy is transferred from the negative energy is transferred from the systemsystem

Work done by a constant forceWork done by a constant force

bull Work done on an object by a constant force Work done on an object by a constant force is defined to be the product of the magnitude of is defined to be the product of the magnitude of the the displacementdisplacement and the component of the and the component of the forceforce parallel to the displacement parallel to the displacement

bull WhereWhere FFIIII is the component of the forceis the component of the force FF parallel to the displacementparallel to the displacement dd

W = FW = FII II middotmiddot d d

WorkIn other words - In other words -

WhereWhere θθ is the angle betweenis the angle between FF and and dd

IfIf θθ is is gt 90gt 90oo work is negative A decelerating car has work is negative A decelerating car has negative work done on it by its enginenegative work done on it by its engine

The unit of work is called The unit of work is called Joule (J)Joule (J) 1J = 1 N1J = 1 Nmiddotmiddotmm

1J=kg∙m1J=kg∙m22ss22

d

F

W = F d cosW = F d cos

Scalar Product of Two VectorsScalar Product of Two Vectors

The scalar product The scalar product of two vectors is of two vectors is written aswritten as A A BB

It is also called It is also called the dot productthe dot product

A A bull bull B = A B cos B = A B cos θθ

θθ is the angle is the angle betweenbetween AA andand BB

Scalar ProductScalar Product

The scalar product is commutativeThe scalar product is commutative A A B = B B = B AA

The scalar product obeys the distributive law The scalar product obeys the distributive law of multiplicationof multiplication

A A (B + C) (B + C) == A A B + A B + A CC

Dot Products of Unit VectorsDot Products of Unit Vectors

bull Using component form withUsing component form with A A andand BB

0kjkiji

1kkjjii

zzyyxx

zyx

zyx

BABABABA

kBjBiBB

kAjAiAA

Work - on and byWork - on and by

A person pushes blockA person pushes block 30 m30 m along the along the ground by exerting force ofground by exerting force of 25 N25 N on the on the trolley How much work does the person do trolley How much work does the person do on the trolleyon the trolley

W = FW = Fd = 25N x 30m = 750 Nmd = 25N x 30m = 750 Nm

Trolley does -Trolley does -750 Nm750 Nm work on the personwork on the person

FFpptFFtrtr

d

A force acts on an object as A force acts on an object as the object moves in the the object moves in the xx direction from the origin direction from the origin to to xx = 500 m = 500 m Find the work Find the work W W done on the done on the object by the forceobject by the force

F 4x ˆ i 3yˆ j N

Mechanical EnergyMechanical Energy

Mechanical energy (energy associated Mechanical energy (energy associated

with with masses) can be thought of as having masses) can be thought of as having

two two components components kinetickinetic and and potentialpotential

ndashKinetic energy is energy of motionKinetic energy is energy of motion

ndashPotential energy is energy of positionPotential energy is energy of position

Kinetic EnergyKinetic Energy

bull In many situations energy can be In many situations energy can be considered asconsidered as ldquothe ability to do workrdquoldquothe ability to do workrdquo

bull Energy can exist in different formsEnergy can exist in different forms

bull Kinetic energy is the energy associated Kinetic energy is the energy associated with the with the motionmotion of an object of an object

bull A moving object can do work on A moving object can do work on another objectanother object

Eg hammer does work on nailEg hammer does work on nail

Kinetic EnergyKinetic Energy Consider an object with massConsider an object with mass mm moving in a straight line with moving in a straight line with initial velocityinitial velocity vvii To accelerate it uniformly to a speedTo accelerate it uniformly to a speed vvff a a constant net forceconstant net force FF is exerted on it parallel to motion over a is exerted on it parallel to motion over a distancedistance dd

Work done on objectWork done on object

W = F d = m a dW = F d = m a d (NII)(NII)SoSo

da

ada

if

iifif

2

2)(222

222

vv

vxxvv

dd

mW if

2

22 vv

Kinetic EnergyKinetic Energy

If we rearrange this we obtainIf we rearrange this we obtain

We define the quantityWe define the quantity frac12mvfrac12mv22 to be the to be the translational translational kinetic energykinetic energy (KE) of the object (KE) of the object

This is the This is the lsquolsquoWork-Energy TheoremWork-Energy Theoremrsquorsquo

ldquoldquoThe net The net workwork done on an object is equal to done on an object is equal to its its change in kinetic energychange in kinetic energyrdquordquo

22

2

1

2

1if mmW vv

W = KE

The Work-Energy TheoremThe Work-Energy Theorem

bull NoteNotendash The net work done on an object is the work The net work done on an object is the work

done by the net forcedone by the net forcendash Units of energy and work must be the Units of energy and work must be the

same (same (JJ))

W = W = KEKE

II Kinetic energyKinetic energy

IIII WorkWork

IIIIII Work - Kinetic energy theoremWork - Kinetic energy theorem

IVIV Work done by a constant forceWork done by a constant force

- Gravitational force- Gravitational force

V Work done by a variable forceV Work done by a variable force

- Spring force- Spring force

1D-Analysis1D-Analysis

3D-Analysis3D-Analysis

VI PowerVI Power

EnergyEnergy scalar quantity associated with a state scalar quantity associated with a state (or condition) of one or more objects (or condition) of one or more objects

II Kinetic energyKinetic energy

Energy associated with the state of motion of an objectEnergy associated with the state of motion of an object

2

2

1mvK

UnitsUnits 1 Joule = 1J = 1 kgm1 Joule = 1J = 1 kgm22ss22

II WorkII Work

Energy transferred ldquotordquo or ldquofromrdquo an Energy transferred ldquotordquo or ldquofromrdquo an object by means of a force acting on the object by means of a force acting on the object object

To To +W+WFrom From -W-W

- - Constant Constant forceforce

xx maF

dFWdFKKvvm

dvvmmaF

d

vvadavv

xxiff

fxxf

xxf

)(2

1

1)(

2

1

22

20

2

20

220

220

2

Work done by the force = Energy transfer due to the forceWork done by the force = Energy transfer due to the force

-To calculate the work done on an object by a force during To calculate the work done on an object by a force during a displacement we use only the force component alonga displacement we use only the force component along the objectrsquos displacement The force component the objectrsquos displacement The force component perpendicular to the displacement does zero workperpendicular to the displacement does zero work

dFFddFW x

cos

Assumptions Assumptions 1) F = constant force 1) F = constant force 2) Object is particle-like (rigid object all 2) Object is particle-like (rigid object all parts of the object must move together)parts of the object must move together)

090

90180

90

W

W

A force doesA force does +W+W when it has a vector component in the when it has a vector component in the same direction as the displacement andsame direction as the displacement and ndashWndashW when it has when it has a vector component in the opposite directiona vector component in the opposite direction W=0W=0 when it when it has no such vector componenthas no such vector component

Net work done by several forces = Sum of works done by Net work done by several forces = Sum of works done by individual forcesindividual forces CalculationCalculation 1) W1) Wnetnet= W= W11+W+W22+W+W33+hellip+hellip

2) F2) Fnet net W Wnetnet=F=Fnetnet d d

II Work-Kinetic Energy TheoremII Work-Kinetic Energy Theorem

WKKK if

Change in the kinetic energy of the Change in the kinetic energy of the particle particle == Net work done on the particle Net work done on the particle

III Work done by a constant forceIII Work done by a constant force

- - Gravitational forceGravitational force

cosmgdW

Rising objectRising object W= mgd cos180W= mgd cos180ordm ordm = -mgd= -mgd FFgg transfers transfers mgdmgd energyenergy from from the objectrsquos kinetic energy the objectrsquos kinetic energy

Falling down objectFalling down object W= mgd cos 0W= mgd cos 0ordm = +mgdordm = +mgd Fg transfers transfers mgd energyenergy to to the objectrsquos kinetic energy the objectrsquos kinetic energy

External applied force + Gravitational force External applied force + Gravitational force

gaif WWKKK

Object stationary before Object stationary before and after the liftand after the lift Wa+Wg=0

The applied force transfers the same The applied force transfers the same amount of energy to the object as the amount of energy to the object as the gravitational force transfers from the objectgravitational force transfers from the object

IV Work done by a variable forceIV Work done by a variable force

- - Spring forceSpring force

xkF

Hookersquos lawHookersquos law

k = spring constant = spring constant measures springrsquos stiffness measures springrsquos stiffness Units Nm

Hookersquos LawHookersquos Law

When When xx is positive is positive (spring is stretched)(spring is stretched) FF is negativeis negative

WhenWhen xx is is 00 (at the (at the equilibrium position)equilibrium position) F F isis 00

When When xx is negative is negative (spring is compressed)(spring is compressed) FF is positiveis positive

Hookersquos LawHookersquos Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the restoring forceis called the restoring force If the block is released it will If the block is released it will

oscillate back and forth between oscillate back and forth between ndashndashxx and and xx

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

Introduction to EnergyIntroduction to Energy

The concept of energy is one of the The concept of energy is one of the most important topics in sciencemost important topics in science

Every physical process that occurs in Every physical process that occurs in the Universe involves energy and the Universe involves energy and energy transfers or transformationsenergy transfers or transformations

Work and EnergyWork and Energy

EnergyEnergy scalar quantity associated with a state (or scalar quantity associated with a state (or condition) of one or more objects condition) of one or more objects

Work and energy are scalars measured in NWork and energy are scalars measured in Nmiddotmiddotm or m or

Joules J 1J=kg∙mJoules J 1J=kg∙m22ss22

Energy can exist in many forms - mechanical Energy can exist in many forms - mechanical

electrical nuclear thermal chemicalhellipelectrical nuclear thermal chemicalhellip

Energy Approach to ProblemsEnergy Approach to Problems

The energy approach to describing The energy approach to describing motion is particularly useful when the motion is particularly useful when the force is not constantforce is not constant

An approach will involve An approach will involve Conservation Conservation of Energyof Energybull This could be extended to biological This could be extended to biological

organisms technological systems and organisms technological systems and engineering situationsengineering situations

Work and EnergyWork and Energy

Energy is a conserved quantity - the total Energy is a conserved quantity - the total amount of energy in the universe is constantamount of energy in the universe is constant

Energy can be converted from one type to Energy can be converted from one type to

another but never destroyedanother but never destroyed

Work and energy concepts can simplify Work and energy concepts can simplify

solutions of mechanical problems - they can solutions of mechanical problems - they can

be used in an alternative analysisbe used in an alternative analysis

SystemsSystems

A A systemsystem is a small portion of the is a small portion of the UniverseUniversebull We will ignore the details of the rest of the We will ignore the details of the rest of the

UniverseUniverse

A critical skill is to identify the systemA critical skill is to identify the system

Valid SystemValid System

A valid system mayA valid system maybull be a single object or particlebe a single object or particlebull be a collection of objects or particlesbe a collection of objects or particlesbull be a region of spacebe a region of spacebull vary in size and shapevary in size and shape

EnvironmentEnvironment

There is a There is a system boundarysystem boundary around the around the systemsystembull The boundary is an imaginary surfaceThe boundary is an imaginary surfacebull It does not necessarily correspond to a physical It does not necessarily correspond to a physical

boundaryboundary

The boundary divides the system from the The boundary divides the system from the environmentenvironmentbull The environment is the rest of the UniverseThe environment is the rest of the Universe

WorkWork

The workThe work WW done on a system by an done on a system by an agent exerting a constant force on the agent exerting a constant force on the system is the product of the magnitudesystem is the product of the magnitude FF of the of the forceforce the magnitude the magnitude ΔΔrr of the of the displacementdisplacement of the point of application of of the point of application of the force andthe force and cos cos θθ where where θθ is the angle is the angle between the force and the displacement between the force and the displacement vectorsvectors

WorkWork

W = F W = F ΔΔr cos r cos θθbull The displacement is that of the point of The displacement is that of the point of

application of the forceapplication of the forcebull A force does no work on the object if the A force does no work on the object if the

object does not move through a displacementobject does not move through a displacementbull The work done by a force on a moving object The work done by a force on a moving object

is zero when the force applied is is zero when the force applied is perpendicular to the displacement of its point perpendicular to the displacement of its point of applicationof application

Work ExampleWork Example

The normal forceThe normal force nn and and the gravitational forcethe gravitational force mgmg do no work on the objectdo no work on the object

cos cos θθ = cos 90deg= 0= cos 90deg= 0

The forceThe force FF does do work does do work on the objecton the object

More About WorkMore About Work The system and the environment must be The system and the environment must be

determined when dealing with workdetermined when dealing with work The environment does work on the systemThe environment does work on the system

(Work (Work byby the environment the environment onon the system) the system) The sign of the work depends on the direction ofThe sign of the work depends on the direction of

FF relative torelative to ΔΔrr Work is positive when projection ofWork is positive when projection of FF ontoonto ΔΔrr

is in the same direction as the displacementis in the same direction as the displacement Work is negative when the projection is in the Work is negative when the projection is in the

opposite directionopposite direction

Work Is An Energy TransferWork Is An Energy Transfer

bull This is important for a system approach This is important for a system approach to solving a problemto solving a problem

bull If the work is done on a system and it is If the work is done on a system and it is positive energy is transferred to the positive energy is transferred to the systemsystem

bull If the work done on the system is If the work done on the system is negative energy is transferred from the negative energy is transferred from the systemsystem

Work done by a constant forceWork done by a constant force

bull Work done on an object by a constant force Work done on an object by a constant force is defined to be the product of the magnitude of is defined to be the product of the magnitude of the the displacementdisplacement and the component of the and the component of the forceforce parallel to the displacement parallel to the displacement

bull WhereWhere FFIIII is the component of the forceis the component of the force FF parallel to the displacementparallel to the displacement dd

W = FW = FII II middotmiddot d d

WorkIn other words - In other words -

WhereWhere θθ is the angle betweenis the angle between FF and and dd

IfIf θθ is is gt 90gt 90oo work is negative A decelerating car has work is negative A decelerating car has negative work done on it by its enginenegative work done on it by its engine

The unit of work is called The unit of work is called Joule (J)Joule (J) 1J = 1 N1J = 1 Nmiddotmiddotmm

1J=kg∙m1J=kg∙m22ss22

d

F

W = F d cosW = F d cos

Scalar Product of Two VectorsScalar Product of Two Vectors

The scalar product The scalar product of two vectors is of two vectors is written aswritten as A A BB

It is also called It is also called the dot productthe dot product

A A bull bull B = A B cos B = A B cos θθ

θθ is the angle is the angle betweenbetween AA andand BB

Scalar ProductScalar Product

The scalar product is commutativeThe scalar product is commutative A A B = B B = B AA

The scalar product obeys the distributive law The scalar product obeys the distributive law of multiplicationof multiplication

A A (B + C) (B + C) == A A B + A B + A CC

Dot Products of Unit VectorsDot Products of Unit Vectors

bull Using component form withUsing component form with A A andand BB

0kjkiji

1kkjjii

zzyyxx

zyx

zyx

BABABABA

kBjBiBB

kAjAiAA

Work - on and byWork - on and by

A person pushes blockA person pushes block 30 m30 m along the along the ground by exerting force ofground by exerting force of 25 N25 N on the on the trolley How much work does the person do trolley How much work does the person do on the trolleyon the trolley

W = FW = Fd = 25N x 30m = 750 Nmd = 25N x 30m = 750 Nm

Trolley does -Trolley does -750 Nm750 Nm work on the personwork on the person

FFpptFFtrtr

d

A force acts on an object as A force acts on an object as the object moves in the the object moves in the xx direction from the origin direction from the origin to to xx = 500 m = 500 m Find the work Find the work W W done on the done on the object by the forceobject by the force

F 4x ˆ i 3yˆ j N

Mechanical EnergyMechanical Energy

Mechanical energy (energy associated Mechanical energy (energy associated

with with masses) can be thought of as having masses) can be thought of as having

two two components components kinetickinetic and and potentialpotential

ndashKinetic energy is energy of motionKinetic energy is energy of motion

ndashPotential energy is energy of positionPotential energy is energy of position

Kinetic EnergyKinetic Energy

bull In many situations energy can be In many situations energy can be considered asconsidered as ldquothe ability to do workrdquoldquothe ability to do workrdquo

bull Energy can exist in different formsEnergy can exist in different forms

bull Kinetic energy is the energy associated Kinetic energy is the energy associated with the with the motionmotion of an object of an object

bull A moving object can do work on A moving object can do work on another objectanother object

Eg hammer does work on nailEg hammer does work on nail

Kinetic EnergyKinetic Energy Consider an object with massConsider an object with mass mm moving in a straight line with moving in a straight line with initial velocityinitial velocity vvii To accelerate it uniformly to a speedTo accelerate it uniformly to a speed vvff a a constant net forceconstant net force FF is exerted on it parallel to motion over a is exerted on it parallel to motion over a distancedistance dd

Work done on objectWork done on object

W = F d = m a dW = F d = m a d (NII)(NII)SoSo

da

ada

if

iifif

2

2)(222

222

vv

vxxvv

dd

mW if

2

22 vv

Kinetic EnergyKinetic Energy

If we rearrange this we obtainIf we rearrange this we obtain

We define the quantityWe define the quantity frac12mvfrac12mv22 to be the to be the translational translational kinetic energykinetic energy (KE) of the object (KE) of the object

This is the This is the lsquolsquoWork-Energy TheoremWork-Energy Theoremrsquorsquo

ldquoldquoThe net The net workwork done on an object is equal to done on an object is equal to its its change in kinetic energychange in kinetic energyrdquordquo

22

2

1

2

1if mmW vv

W = KE

The Work-Energy TheoremThe Work-Energy Theorem

bull NoteNotendash The net work done on an object is the work The net work done on an object is the work

done by the net forcedone by the net forcendash Units of energy and work must be the Units of energy and work must be the

same (same (JJ))

W = W = KEKE

II Kinetic energyKinetic energy

IIII WorkWork

IIIIII Work - Kinetic energy theoremWork - Kinetic energy theorem

IVIV Work done by a constant forceWork done by a constant force

- Gravitational force- Gravitational force

V Work done by a variable forceV Work done by a variable force

- Spring force- Spring force

1D-Analysis1D-Analysis

3D-Analysis3D-Analysis

VI PowerVI Power

EnergyEnergy scalar quantity associated with a state scalar quantity associated with a state (or condition) of one or more objects (or condition) of one or more objects

II Kinetic energyKinetic energy

Energy associated with the state of motion of an objectEnergy associated with the state of motion of an object

2

2

1mvK

UnitsUnits 1 Joule = 1J = 1 kgm1 Joule = 1J = 1 kgm22ss22

II WorkII Work

Energy transferred ldquotordquo or ldquofromrdquo an Energy transferred ldquotordquo or ldquofromrdquo an object by means of a force acting on the object by means of a force acting on the object object

To To +W+WFrom From -W-W

- - Constant Constant forceforce

xx maF

dFWdFKKvvm

dvvmmaF

d

vvadavv

xxiff

fxxf

xxf

)(2

1

1)(

2

1

22

20

2

20

220

220

2

Work done by the force = Energy transfer due to the forceWork done by the force = Energy transfer due to the force

-To calculate the work done on an object by a force during To calculate the work done on an object by a force during a displacement we use only the force component alonga displacement we use only the force component along the objectrsquos displacement The force component the objectrsquos displacement The force component perpendicular to the displacement does zero workperpendicular to the displacement does zero work

dFFddFW x

cos

Assumptions Assumptions 1) F = constant force 1) F = constant force 2) Object is particle-like (rigid object all 2) Object is particle-like (rigid object all parts of the object must move together)parts of the object must move together)

090

90180

90

W

W

A force doesA force does +W+W when it has a vector component in the when it has a vector component in the same direction as the displacement andsame direction as the displacement and ndashWndashW when it has when it has a vector component in the opposite directiona vector component in the opposite direction W=0W=0 when it when it has no such vector componenthas no such vector component

Net work done by several forces = Sum of works done by Net work done by several forces = Sum of works done by individual forcesindividual forces CalculationCalculation 1) W1) Wnetnet= W= W11+W+W22+W+W33+hellip+hellip

2) F2) Fnet net W Wnetnet=F=Fnetnet d d

II Work-Kinetic Energy TheoremII Work-Kinetic Energy Theorem

WKKK if

Change in the kinetic energy of the Change in the kinetic energy of the particle particle == Net work done on the particle Net work done on the particle

III Work done by a constant forceIII Work done by a constant force

- - Gravitational forceGravitational force

cosmgdW

Rising objectRising object W= mgd cos180W= mgd cos180ordm ordm = -mgd= -mgd FFgg transfers transfers mgdmgd energyenergy from from the objectrsquos kinetic energy the objectrsquos kinetic energy

Falling down objectFalling down object W= mgd cos 0W= mgd cos 0ordm = +mgdordm = +mgd Fg transfers transfers mgd energyenergy to to the objectrsquos kinetic energy the objectrsquos kinetic energy

External applied force + Gravitational force External applied force + Gravitational force

gaif WWKKK

Object stationary before Object stationary before and after the liftand after the lift Wa+Wg=0

The applied force transfers the same The applied force transfers the same amount of energy to the object as the amount of energy to the object as the gravitational force transfers from the objectgravitational force transfers from the object

IV Work done by a variable forceIV Work done by a variable force

- - Spring forceSpring force

xkF

Hookersquos lawHookersquos law

k = spring constant = spring constant measures springrsquos stiffness measures springrsquos stiffness Units Nm

Hookersquos LawHookersquos Law

When When xx is positive is positive (spring is stretched)(spring is stretched) FF is negativeis negative

WhenWhen xx is is 00 (at the (at the equilibrium position)equilibrium position) F F isis 00

When When xx is negative is negative (spring is compressed)(spring is compressed) FF is positiveis positive

Hookersquos LawHookersquos Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the restoring forceis called the restoring force If the block is released it will If the block is released it will

oscillate back and forth between oscillate back and forth between ndashndashxx and and xx

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

Work and EnergyWork and Energy

EnergyEnergy scalar quantity associated with a state (or scalar quantity associated with a state (or condition) of one or more objects condition) of one or more objects

Work and energy are scalars measured in NWork and energy are scalars measured in Nmiddotmiddotm or m or

Joules J 1J=kg∙mJoules J 1J=kg∙m22ss22

Energy can exist in many forms - mechanical Energy can exist in many forms - mechanical

electrical nuclear thermal chemicalhellipelectrical nuclear thermal chemicalhellip

Energy Approach to ProblemsEnergy Approach to Problems

The energy approach to describing The energy approach to describing motion is particularly useful when the motion is particularly useful when the force is not constantforce is not constant

An approach will involve An approach will involve Conservation Conservation of Energyof Energybull This could be extended to biological This could be extended to biological

organisms technological systems and organisms technological systems and engineering situationsengineering situations

Work and EnergyWork and Energy

Energy is a conserved quantity - the total Energy is a conserved quantity - the total amount of energy in the universe is constantamount of energy in the universe is constant

Energy can be converted from one type to Energy can be converted from one type to

another but never destroyedanother but never destroyed

Work and energy concepts can simplify Work and energy concepts can simplify

solutions of mechanical problems - they can solutions of mechanical problems - they can

be used in an alternative analysisbe used in an alternative analysis

SystemsSystems

A A systemsystem is a small portion of the is a small portion of the UniverseUniversebull We will ignore the details of the rest of the We will ignore the details of the rest of the

UniverseUniverse

A critical skill is to identify the systemA critical skill is to identify the system

Valid SystemValid System

A valid system mayA valid system maybull be a single object or particlebe a single object or particlebull be a collection of objects or particlesbe a collection of objects or particlesbull be a region of spacebe a region of spacebull vary in size and shapevary in size and shape

EnvironmentEnvironment

There is a There is a system boundarysystem boundary around the around the systemsystembull The boundary is an imaginary surfaceThe boundary is an imaginary surfacebull It does not necessarily correspond to a physical It does not necessarily correspond to a physical

boundaryboundary

The boundary divides the system from the The boundary divides the system from the environmentenvironmentbull The environment is the rest of the UniverseThe environment is the rest of the Universe

WorkWork

The workThe work WW done on a system by an done on a system by an agent exerting a constant force on the agent exerting a constant force on the system is the product of the magnitudesystem is the product of the magnitude FF of the of the forceforce the magnitude the magnitude ΔΔrr of the of the displacementdisplacement of the point of application of of the point of application of the force andthe force and cos cos θθ where where θθ is the angle is the angle between the force and the displacement between the force and the displacement vectorsvectors

WorkWork

W = F W = F ΔΔr cos r cos θθbull The displacement is that of the point of The displacement is that of the point of

application of the forceapplication of the forcebull A force does no work on the object if the A force does no work on the object if the

object does not move through a displacementobject does not move through a displacementbull The work done by a force on a moving object The work done by a force on a moving object

is zero when the force applied is is zero when the force applied is perpendicular to the displacement of its point perpendicular to the displacement of its point of applicationof application

Work ExampleWork Example

The normal forceThe normal force nn and and the gravitational forcethe gravitational force mgmg do no work on the objectdo no work on the object

cos cos θθ = cos 90deg= 0= cos 90deg= 0

The forceThe force FF does do work does do work on the objecton the object

More About WorkMore About Work The system and the environment must be The system and the environment must be

determined when dealing with workdetermined when dealing with work The environment does work on the systemThe environment does work on the system

(Work (Work byby the environment the environment onon the system) the system) The sign of the work depends on the direction ofThe sign of the work depends on the direction of

FF relative torelative to ΔΔrr Work is positive when projection ofWork is positive when projection of FF ontoonto ΔΔrr

is in the same direction as the displacementis in the same direction as the displacement Work is negative when the projection is in the Work is negative when the projection is in the

opposite directionopposite direction

Work Is An Energy TransferWork Is An Energy Transfer

bull This is important for a system approach This is important for a system approach to solving a problemto solving a problem

bull If the work is done on a system and it is If the work is done on a system and it is positive energy is transferred to the positive energy is transferred to the systemsystem

bull If the work done on the system is If the work done on the system is negative energy is transferred from the negative energy is transferred from the systemsystem

Work done by a constant forceWork done by a constant force

bull Work done on an object by a constant force Work done on an object by a constant force is defined to be the product of the magnitude of is defined to be the product of the magnitude of the the displacementdisplacement and the component of the and the component of the forceforce parallel to the displacement parallel to the displacement

bull WhereWhere FFIIII is the component of the forceis the component of the force FF parallel to the displacementparallel to the displacement dd

W = FW = FII II middotmiddot d d

WorkIn other words - In other words -

WhereWhere θθ is the angle betweenis the angle between FF and and dd

IfIf θθ is is gt 90gt 90oo work is negative A decelerating car has work is negative A decelerating car has negative work done on it by its enginenegative work done on it by its engine

The unit of work is called The unit of work is called Joule (J)Joule (J) 1J = 1 N1J = 1 Nmiddotmiddotmm

1J=kg∙m1J=kg∙m22ss22

d

F

W = F d cosW = F d cos

Scalar Product of Two VectorsScalar Product of Two Vectors

The scalar product The scalar product of two vectors is of two vectors is written aswritten as A A BB

It is also called It is also called the dot productthe dot product

A A bull bull B = A B cos B = A B cos θθ

θθ is the angle is the angle betweenbetween AA andand BB

Scalar ProductScalar Product

The scalar product is commutativeThe scalar product is commutative A A B = B B = B AA

The scalar product obeys the distributive law The scalar product obeys the distributive law of multiplicationof multiplication

A A (B + C) (B + C) == A A B + A B + A CC

Dot Products of Unit VectorsDot Products of Unit Vectors

bull Using component form withUsing component form with A A andand BB

0kjkiji

1kkjjii

zzyyxx

zyx

zyx

BABABABA

kBjBiBB

kAjAiAA

Work - on and byWork - on and by

A person pushes blockA person pushes block 30 m30 m along the along the ground by exerting force ofground by exerting force of 25 N25 N on the on the trolley How much work does the person do trolley How much work does the person do on the trolleyon the trolley

W = FW = Fd = 25N x 30m = 750 Nmd = 25N x 30m = 750 Nm

Trolley does -Trolley does -750 Nm750 Nm work on the personwork on the person

FFpptFFtrtr

d

A force acts on an object as A force acts on an object as the object moves in the the object moves in the xx direction from the origin direction from the origin to to xx = 500 m = 500 m Find the work Find the work W W done on the done on the object by the forceobject by the force

F 4x ˆ i 3yˆ j N

Mechanical EnergyMechanical Energy

Mechanical energy (energy associated Mechanical energy (energy associated

with with masses) can be thought of as having masses) can be thought of as having

two two components components kinetickinetic and and potentialpotential

ndashKinetic energy is energy of motionKinetic energy is energy of motion

ndashPotential energy is energy of positionPotential energy is energy of position

Kinetic EnergyKinetic Energy

bull In many situations energy can be In many situations energy can be considered asconsidered as ldquothe ability to do workrdquoldquothe ability to do workrdquo

bull Energy can exist in different formsEnergy can exist in different forms

bull Kinetic energy is the energy associated Kinetic energy is the energy associated with the with the motionmotion of an object of an object

bull A moving object can do work on A moving object can do work on another objectanother object

Eg hammer does work on nailEg hammer does work on nail

Kinetic EnergyKinetic Energy Consider an object with massConsider an object with mass mm moving in a straight line with moving in a straight line with initial velocityinitial velocity vvii To accelerate it uniformly to a speedTo accelerate it uniformly to a speed vvff a a constant net forceconstant net force FF is exerted on it parallel to motion over a is exerted on it parallel to motion over a distancedistance dd

Work done on objectWork done on object

W = F d = m a dW = F d = m a d (NII)(NII)SoSo

da

ada

if

iifif

2

2)(222

222

vv

vxxvv

dd

mW if

2

22 vv

Kinetic EnergyKinetic Energy

If we rearrange this we obtainIf we rearrange this we obtain

We define the quantityWe define the quantity frac12mvfrac12mv22 to be the to be the translational translational kinetic energykinetic energy (KE) of the object (KE) of the object

This is the This is the lsquolsquoWork-Energy TheoremWork-Energy Theoremrsquorsquo

ldquoldquoThe net The net workwork done on an object is equal to done on an object is equal to its its change in kinetic energychange in kinetic energyrdquordquo

22

2

1

2

1if mmW vv

W = KE

The Work-Energy TheoremThe Work-Energy Theorem

bull NoteNotendash The net work done on an object is the work The net work done on an object is the work

done by the net forcedone by the net forcendash Units of energy and work must be the Units of energy and work must be the

same (same (JJ))

W = W = KEKE

II Kinetic energyKinetic energy

IIII WorkWork

IIIIII Work - Kinetic energy theoremWork - Kinetic energy theorem

IVIV Work done by a constant forceWork done by a constant force

- Gravitational force- Gravitational force

V Work done by a variable forceV Work done by a variable force

- Spring force- Spring force

1D-Analysis1D-Analysis

3D-Analysis3D-Analysis

VI PowerVI Power

EnergyEnergy scalar quantity associated with a state scalar quantity associated with a state (or condition) of one or more objects (or condition) of one or more objects

II Kinetic energyKinetic energy

Energy associated with the state of motion of an objectEnergy associated with the state of motion of an object

2

2

1mvK

UnitsUnits 1 Joule = 1J = 1 kgm1 Joule = 1J = 1 kgm22ss22

II WorkII Work

Energy transferred ldquotordquo or ldquofromrdquo an Energy transferred ldquotordquo or ldquofromrdquo an object by means of a force acting on the object by means of a force acting on the object object

To To +W+WFrom From -W-W

- - Constant Constant forceforce

xx maF

dFWdFKKvvm

dvvmmaF

d

vvadavv

xxiff

fxxf

xxf

)(2

1

1)(

2

1

22

20

2

20

220

220

2

Work done by the force = Energy transfer due to the forceWork done by the force = Energy transfer due to the force

-To calculate the work done on an object by a force during To calculate the work done on an object by a force during a displacement we use only the force component alonga displacement we use only the force component along the objectrsquos displacement The force component the objectrsquos displacement The force component perpendicular to the displacement does zero workperpendicular to the displacement does zero work

dFFddFW x

cos

Assumptions Assumptions 1) F = constant force 1) F = constant force 2) Object is particle-like (rigid object all 2) Object is particle-like (rigid object all parts of the object must move together)parts of the object must move together)

090

90180

90

W

W

A force doesA force does +W+W when it has a vector component in the when it has a vector component in the same direction as the displacement andsame direction as the displacement and ndashWndashW when it has when it has a vector component in the opposite directiona vector component in the opposite direction W=0W=0 when it when it has no such vector componenthas no such vector component

Net work done by several forces = Sum of works done by Net work done by several forces = Sum of works done by individual forcesindividual forces CalculationCalculation 1) W1) Wnetnet= W= W11+W+W22+W+W33+hellip+hellip

2) F2) Fnet net W Wnetnet=F=Fnetnet d d

II Work-Kinetic Energy TheoremII Work-Kinetic Energy Theorem

WKKK if

Change in the kinetic energy of the Change in the kinetic energy of the particle particle == Net work done on the particle Net work done on the particle

III Work done by a constant forceIII Work done by a constant force

- - Gravitational forceGravitational force

cosmgdW

Rising objectRising object W= mgd cos180W= mgd cos180ordm ordm = -mgd= -mgd FFgg transfers transfers mgdmgd energyenergy from from the objectrsquos kinetic energy the objectrsquos kinetic energy

Falling down objectFalling down object W= mgd cos 0W= mgd cos 0ordm = +mgdordm = +mgd Fg transfers transfers mgd energyenergy to to the objectrsquos kinetic energy the objectrsquos kinetic energy

External applied force + Gravitational force External applied force + Gravitational force

gaif WWKKK

Object stationary before Object stationary before and after the liftand after the lift Wa+Wg=0

The applied force transfers the same The applied force transfers the same amount of energy to the object as the amount of energy to the object as the gravitational force transfers from the objectgravitational force transfers from the object

IV Work done by a variable forceIV Work done by a variable force

- - Spring forceSpring force

xkF

Hookersquos lawHookersquos law

k = spring constant = spring constant measures springrsquos stiffness measures springrsquos stiffness Units Nm

Hookersquos LawHookersquos Law

When When xx is positive is positive (spring is stretched)(spring is stretched) FF is negativeis negative

WhenWhen xx is is 00 (at the (at the equilibrium position)equilibrium position) F F isis 00

When When xx is negative is negative (spring is compressed)(spring is compressed) FF is positiveis positive

Hookersquos LawHookersquos Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the restoring forceis called the restoring force If the block is released it will If the block is released it will

oscillate back and forth between oscillate back and forth between ndashndashxx and and xx

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

Energy Approach to ProblemsEnergy Approach to Problems

The energy approach to describing The energy approach to describing motion is particularly useful when the motion is particularly useful when the force is not constantforce is not constant

An approach will involve An approach will involve Conservation Conservation of Energyof Energybull This could be extended to biological This could be extended to biological

organisms technological systems and organisms technological systems and engineering situationsengineering situations

Work and EnergyWork and Energy

Energy is a conserved quantity - the total Energy is a conserved quantity - the total amount of energy in the universe is constantamount of energy in the universe is constant

Energy can be converted from one type to Energy can be converted from one type to

another but never destroyedanother but never destroyed

Work and energy concepts can simplify Work and energy concepts can simplify

solutions of mechanical problems - they can solutions of mechanical problems - they can

be used in an alternative analysisbe used in an alternative analysis

SystemsSystems

A A systemsystem is a small portion of the is a small portion of the UniverseUniversebull We will ignore the details of the rest of the We will ignore the details of the rest of the

UniverseUniverse

A critical skill is to identify the systemA critical skill is to identify the system

Valid SystemValid System

A valid system mayA valid system maybull be a single object or particlebe a single object or particlebull be a collection of objects or particlesbe a collection of objects or particlesbull be a region of spacebe a region of spacebull vary in size and shapevary in size and shape

EnvironmentEnvironment

There is a There is a system boundarysystem boundary around the around the systemsystembull The boundary is an imaginary surfaceThe boundary is an imaginary surfacebull It does not necessarily correspond to a physical It does not necessarily correspond to a physical

boundaryboundary

The boundary divides the system from the The boundary divides the system from the environmentenvironmentbull The environment is the rest of the UniverseThe environment is the rest of the Universe

WorkWork

The workThe work WW done on a system by an done on a system by an agent exerting a constant force on the agent exerting a constant force on the system is the product of the magnitudesystem is the product of the magnitude FF of the of the forceforce the magnitude the magnitude ΔΔrr of the of the displacementdisplacement of the point of application of of the point of application of the force andthe force and cos cos θθ where where θθ is the angle is the angle between the force and the displacement between the force and the displacement vectorsvectors

WorkWork

W = F W = F ΔΔr cos r cos θθbull The displacement is that of the point of The displacement is that of the point of

application of the forceapplication of the forcebull A force does no work on the object if the A force does no work on the object if the

object does not move through a displacementobject does not move through a displacementbull The work done by a force on a moving object The work done by a force on a moving object

is zero when the force applied is is zero when the force applied is perpendicular to the displacement of its point perpendicular to the displacement of its point of applicationof application

Work ExampleWork Example

The normal forceThe normal force nn and and the gravitational forcethe gravitational force mgmg do no work on the objectdo no work on the object

cos cos θθ = cos 90deg= 0= cos 90deg= 0

The forceThe force FF does do work does do work on the objecton the object

More About WorkMore About Work The system and the environment must be The system and the environment must be

determined when dealing with workdetermined when dealing with work The environment does work on the systemThe environment does work on the system

(Work (Work byby the environment the environment onon the system) the system) The sign of the work depends on the direction ofThe sign of the work depends on the direction of

FF relative torelative to ΔΔrr Work is positive when projection ofWork is positive when projection of FF ontoonto ΔΔrr

is in the same direction as the displacementis in the same direction as the displacement Work is negative when the projection is in the Work is negative when the projection is in the

opposite directionopposite direction

Work Is An Energy TransferWork Is An Energy Transfer

bull This is important for a system approach This is important for a system approach to solving a problemto solving a problem

bull If the work is done on a system and it is If the work is done on a system and it is positive energy is transferred to the positive energy is transferred to the systemsystem

bull If the work done on the system is If the work done on the system is negative energy is transferred from the negative energy is transferred from the systemsystem

Work done by a constant forceWork done by a constant force

bull Work done on an object by a constant force Work done on an object by a constant force is defined to be the product of the magnitude of is defined to be the product of the magnitude of the the displacementdisplacement and the component of the and the component of the forceforce parallel to the displacement parallel to the displacement

bull WhereWhere FFIIII is the component of the forceis the component of the force FF parallel to the displacementparallel to the displacement dd

W = FW = FII II middotmiddot d d

WorkIn other words - In other words -

WhereWhere θθ is the angle betweenis the angle between FF and and dd

IfIf θθ is is gt 90gt 90oo work is negative A decelerating car has work is negative A decelerating car has negative work done on it by its enginenegative work done on it by its engine

The unit of work is called The unit of work is called Joule (J)Joule (J) 1J = 1 N1J = 1 Nmiddotmiddotmm

1J=kg∙m1J=kg∙m22ss22

d

F

W = F d cosW = F d cos

Scalar Product of Two VectorsScalar Product of Two Vectors

The scalar product The scalar product of two vectors is of two vectors is written aswritten as A A BB

It is also called It is also called the dot productthe dot product

A A bull bull B = A B cos B = A B cos θθ

θθ is the angle is the angle betweenbetween AA andand BB

Scalar ProductScalar Product

The scalar product is commutativeThe scalar product is commutative A A B = B B = B AA

The scalar product obeys the distributive law The scalar product obeys the distributive law of multiplicationof multiplication

A A (B + C) (B + C) == A A B + A B + A CC

Dot Products of Unit VectorsDot Products of Unit Vectors

bull Using component form withUsing component form with A A andand BB

0kjkiji

1kkjjii

zzyyxx

zyx

zyx

BABABABA

kBjBiBB

kAjAiAA

Work - on and byWork - on and by

A person pushes blockA person pushes block 30 m30 m along the along the ground by exerting force ofground by exerting force of 25 N25 N on the on the trolley How much work does the person do trolley How much work does the person do on the trolleyon the trolley

W = FW = Fd = 25N x 30m = 750 Nmd = 25N x 30m = 750 Nm

Trolley does -Trolley does -750 Nm750 Nm work on the personwork on the person

FFpptFFtrtr

d

A force acts on an object as A force acts on an object as the object moves in the the object moves in the xx direction from the origin direction from the origin to to xx = 500 m = 500 m Find the work Find the work W W done on the done on the object by the forceobject by the force

F 4x ˆ i 3yˆ j N

Mechanical EnergyMechanical Energy

Mechanical energy (energy associated Mechanical energy (energy associated

with with masses) can be thought of as having masses) can be thought of as having

two two components components kinetickinetic and and potentialpotential

ndashKinetic energy is energy of motionKinetic energy is energy of motion

ndashPotential energy is energy of positionPotential energy is energy of position

Kinetic EnergyKinetic Energy

bull In many situations energy can be In many situations energy can be considered asconsidered as ldquothe ability to do workrdquoldquothe ability to do workrdquo

bull Energy can exist in different formsEnergy can exist in different forms

bull Kinetic energy is the energy associated Kinetic energy is the energy associated with the with the motionmotion of an object of an object

bull A moving object can do work on A moving object can do work on another objectanother object

Eg hammer does work on nailEg hammer does work on nail

Kinetic EnergyKinetic Energy Consider an object with massConsider an object with mass mm moving in a straight line with moving in a straight line with initial velocityinitial velocity vvii To accelerate it uniformly to a speedTo accelerate it uniformly to a speed vvff a a constant net forceconstant net force FF is exerted on it parallel to motion over a is exerted on it parallel to motion over a distancedistance dd

Work done on objectWork done on object

W = F d = m a dW = F d = m a d (NII)(NII)SoSo

da

ada

if

iifif

2

2)(222

222

vv

vxxvv

dd

mW if

2

22 vv

Kinetic EnergyKinetic Energy

If we rearrange this we obtainIf we rearrange this we obtain

We define the quantityWe define the quantity frac12mvfrac12mv22 to be the to be the translational translational kinetic energykinetic energy (KE) of the object (KE) of the object

This is the This is the lsquolsquoWork-Energy TheoremWork-Energy Theoremrsquorsquo

ldquoldquoThe net The net workwork done on an object is equal to done on an object is equal to its its change in kinetic energychange in kinetic energyrdquordquo

22

2

1

2

1if mmW vv

W = KE

The Work-Energy TheoremThe Work-Energy Theorem

bull NoteNotendash The net work done on an object is the work The net work done on an object is the work

done by the net forcedone by the net forcendash Units of energy and work must be the Units of energy and work must be the

same (same (JJ))

W = W = KEKE

II Kinetic energyKinetic energy

IIII WorkWork

IIIIII Work - Kinetic energy theoremWork - Kinetic energy theorem

IVIV Work done by a constant forceWork done by a constant force

- Gravitational force- Gravitational force

V Work done by a variable forceV Work done by a variable force

- Spring force- Spring force

1D-Analysis1D-Analysis

3D-Analysis3D-Analysis

VI PowerVI Power

EnergyEnergy scalar quantity associated with a state scalar quantity associated with a state (or condition) of one or more objects (or condition) of one or more objects

II Kinetic energyKinetic energy

Energy associated with the state of motion of an objectEnergy associated with the state of motion of an object

2

2

1mvK

UnitsUnits 1 Joule = 1J = 1 kgm1 Joule = 1J = 1 kgm22ss22

II WorkII Work

Energy transferred ldquotordquo or ldquofromrdquo an Energy transferred ldquotordquo or ldquofromrdquo an object by means of a force acting on the object by means of a force acting on the object object

To To +W+WFrom From -W-W

- - Constant Constant forceforce

xx maF

dFWdFKKvvm

dvvmmaF

d

vvadavv

xxiff

fxxf

xxf

)(2

1

1)(

2

1

22

20

2

20

220

220

2

Work done by the force = Energy transfer due to the forceWork done by the force = Energy transfer due to the force

-To calculate the work done on an object by a force during To calculate the work done on an object by a force during a displacement we use only the force component alonga displacement we use only the force component along the objectrsquos displacement The force component the objectrsquos displacement The force component perpendicular to the displacement does zero workperpendicular to the displacement does zero work

dFFddFW x

cos

Assumptions Assumptions 1) F = constant force 1) F = constant force 2) Object is particle-like (rigid object all 2) Object is particle-like (rigid object all parts of the object must move together)parts of the object must move together)

090

90180

90

W

W

A force doesA force does +W+W when it has a vector component in the when it has a vector component in the same direction as the displacement andsame direction as the displacement and ndashWndashW when it has when it has a vector component in the opposite directiona vector component in the opposite direction W=0W=0 when it when it has no such vector componenthas no such vector component

Net work done by several forces = Sum of works done by Net work done by several forces = Sum of works done by individual forcesindividual forces CalculationCalculation 1) W1) Wnetnet= W= W11+W+W22+W+W33+hellip+hellip

2) F2) Fnet net W Wnetnet=F=Fnetnet d d

II Work-Kinetic Energy TheoremII Work-Kinetic Energy Theorem

WKKK if

Change in the kinetic energy of the Change in the kinetic energy of the particle particle == Net work done on the particle Net work done on the particle

III Work done by a constant forceIII Work done by a constant force

- - Gravitational forceGravitational force

cosmgdW

Rising objectRising object W= mgd cos180W= mgd cos180ordm ordm = -mgd= -mgd FFgg transfers transfers mgdmgd energyenergy from from the objectrsquos kinetic energy the objectrsquos kinetic energy

Falling down objectFalling down object W= mgd cos 0W= mgd cos 0ordm = +mgdordm = +mgd Fg transfers transfers mgd energyenergy to to the objectrsquos kinetic energy the objectrsquos kinetic energy

External applied force + Gravitational force External applied force + Gravitational force

gaif WWKKK

Object stationary before Object stationary before and after the liftand after the lift Wa+Wg=0

The applied force transfers the same The applied force transfers the same amount of energy to the object as the amount of energy to the object as the gravitational force transfers from the objectgravitational force transfers from the object

IV Work done by a variable forceIV Work done by a variable force

- - Spring forceSpring force

xkF

Hookersquos lawHookersquos law

k = spring constant = spring constant measures springrsquos stiffness measures springrsquos stiffness Units Nm

Hookersquos LawHookersquos Law

When When xx is positive is positive (spring is stretched)(spring is stretched) FF is negativeis negative

WhenWhen xx is is 00 (at the (at the equilibrium position)equilibrium position) F F isis 00

When When xx is negative is negative (spring is compressed)(spring is compressed) FF is positiveis positive

Hookersquos LawHookersquos Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the restoring forceis called the restoring force If the block is released it will If the block is released it will

oscillate back and forth between oscillate back and forth between ndashndashxx and and xx

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

Work and EnergyWork and Energy

Energy is a conserved quantity - the total Energy is a conserved quantity - the total amount of energy in the universe is constantamount of energy in the universe is constant

Energy can be converted from one type to Energy can be converted from one type to

another but never destroyedanother but never destroyed

Work and energy concepts can simplify Work and energy concepts can simplify

solutions of mechanical problems - they can solutions of mechanical problems - they can

be used in an alternative analysisbe used in an alternative analysis

SystemsSystems

A A systemsystem is a small portion of the is a small portion of the UniverseUniversebull We will ignore the details of the rest of the We will ignore the details of the rest of the

UniverseUniverse

A critical skill is to identify the systemA critical skill is to identify the system

Valid SystemValid System

A valid system mayA valid system maybull be a single object or particlebe a single object or particlebull be a collection of objects or particlesbe a collection of objects or particlesbull be a region of spacebe a region of spacebull vary in size and shapevary in size and shape

EnvironmentEnvironment

There is a There is a system boundarysystem boundary around the around the systemsystembull The boundary is an imaginary surfaceThe boundary is an imaginary surfacebull It does not necessarily correspond to a physical It does not necessarily correspond to a physical

boundaryboundary

The boundary divides the system from the The boundary divides the system from the environmentenvironmentbull The environment is the rest of the UniverseThe environment is the rest of the Universe

WorkWork

The workThe work WW done on a system by an done on a system by an agent exerting a constant force on the agent exerting a constant force on the system is the product of the magnitudesystem is the product of the magnitude FF of the of the forceforce the magnitude the magnitude ΔΔrr of the of the displacementdisplacement of the point of application of of the point of application of the force andthe force and cos cos θθ where where θθ is the angle is the angle between the force and the displacement between the force and the displacement vectorsvectors

WorkWork

W = F W = F ΔΔr cos r cos θθbull The displacement is that of the point of The displacement is that of the point of

application of the forceapplication of the forcebull A force does no work on the object if the A force does no work on the object if the

object does not move through a displacementobject does not move through a displacementbull The work done by a force on a moving object The work done by a force on a moving object

is zero when the force applied is is zero when the force applied is perpendicular to the displacement of its point perpendicular to the displacement of its point of applicationof application

Work ExampleWork Example

The normal forceThe normal force nn and and the gravitational forcethe gravitational force mgmg do no work on the objectdo no work on the object

cos cos θθ = cos 90deg= 0= cos 90deg= 0

The forceThe force FF does do work does do work on the objecton the object

More About WorkMore About Work The system and the environment must be The system and the environment must be

determined when dealing with workdetermined when dealing with work The environment does work on the systemThe environment does work on the system

(Work (Work byby the environment the environment onon the system) the system) The sign of the work depends on the direction ofThe sign of the work depends on the direction of

FF relative torelative to ΔΔrr Work is positive when projection ofWork is positive when projection of FF ontoonto ΔΔrr

is in the same direction as the displacementis in the same direction as the displacement Work is negative when the projection is in the Work is negative when the projection is in the

opposite directionopposite direction

Work Is An Energy TransferWork Is An Energy Transfer

bull This is important for a system approach This is important for a system approach to solving a problemto solving a problem

bull If the work is done on a system and it is If the work is done on a system and it is positive energy is transferred to the positive energy is transferred to the systemsystem

bull If the work done on the system is If the work done on the system is negative energy is transferred from the negative energy is transferred from the systemsystem

Work done by a constant forceWork done by a constant force

bull Work done on an object by a constant force Work done on an object by a constant force is defined to be the product of the magnitude of is defined to be the product of the magnitude of the the displacementdisplacement and the component of the and the component of the forceforce parallel to the displacement parallel to the displacement

bull WhereWhere FFIIII is the component of the forceis the component of the force FF parallel to the displacementparallel to the displacement dd

W = FW = FII II middotmiddot d d

WorkIn other words - In other words -

WhereWhere θθ is the angle betweenis the angle between FF and and dd

IfIf θθ is is gt 90gt 90oo work is negative A decelerating car has work is negative A decelerating car has negative work done on it by its enginenegative work done on it by its engine

The unit of work is called The unit of work is called Joule (J)Joule (J) 1J = 1 N1J = 1 Nmiddotmiddotmm

1J=kg∙m1J=kg∙m22ss22

d

F

W = F d cosW = F d cos

Scalar Product of Two VectorsScalar Product of Two Vectors

The scalar product The scalar product of two vectors is of two vectors is written aswritten as A A BB

It is also called It is also called the dot productthe dot product

A A bull bull B = A B cos B = A B cos θθ

θθ is the angle is the angle betweenbetween AA andand BB

Scalar ProductScalar Product

The scalar product is commutativeThe scalar product is commutative A A B = B B = B AA

The scalar product obeys the distributive law The scalar product obeys the distributive law of multiplicationof multiplication

A A (B + C) (B + C) == A A B + A B + A CC

Dot Products of Unit VectorsDot Products of Unit Vectors

bull Using component form withUsing component form with A A andand BB

0kjkiji

1kkjjii

zzyyxx

zyx

zyx

BABABABA

kBjBiBB

kAjAiAA

Work - on and byWork - on and by

A person pushes blockA person pushes block 30 m30 m along the along the ground by exerting force ofground by exerting force of 25 N25 N on the on the trolley How much work does the person do trolley How much work does the person do on the trolleyon the trolley

W = FW = Fd = 25N x 30m = 750 Nmd = 25N x 30m = 750 Nm

Trolley does -Trolley does -750 Nm750 Nm work on the personwork on the person

FFpptFFtrtr

d

A force acts on an object as A force acts on an object as the object moves in the the object moves in the xx direction from the origin direction from the origin to to xx = 500 m = 500 m Find the work Find the work W W done on the done on the object by the forceobject by the force

F 4x ˆ i 3yˆ j N

Mechanical EnergyMechanical Energy

Mechanical energy (energy associated Mechanical energy (energy associated

with with masses) can be thought of as having masses) can be thought of as having

two two components components kinetickinetic and and potentialpotential

ndashKinetic energy is energy of motionKinetic energy is energy of motion

ndashPotential energy is energy of positionPotential energy is energy of position

Kinetic EnergyKinetic Energy

bull In many situations energy can be In many situations energy can be considered asconsidered as ldquothe ability to do workrdquoldquothe ability to do workrdquo

bull Energy can exist in different formsEnergy can exist in different forms

bull Kinetic energy is the energy associated Kinetic energy is the energy associated with the with the motionmotion of an object of an object

bull A moving object can do work on A moving object can do work on another objectanother object

Eg hammer does work on nailEg hammer does work on nail

Kinetic EnergyKinetic Energy Consider an object with massConsider an object with mass mm moving in a straight line with moving in a straight line with initial velocityinitial velocity vvii To accelerate it uniformly to a speedTo accelerate it uniformly to a speed vvff a a constant net forceconstant net force FF is exerted on it parallel to motion over a is exerted on it parallel to motion over a distancedistance dd

Work done on objectWork done on object

W = F d = m a dW = F d = m a d (NII)(NII)SoSo

da

ada

if

iifif

2

2)(222

222

vv

vxxvv

dd

mW if

2

22 vv

Kinetic EnergyKinetic Energy

If we rearrange this we obtainIf we rearrange this we obtain

We define the quantityWe define the quantity frac12mvfrac12mv22 to be the to be the translational translational kinetic energykinetic energy (KE) of the object (KE) of the object

This is the This is the lsquolsquoWork-Energy TheoremWork-Energy Theoremrsquorsquo

ldquoldquoThe net The net workwork done on an object is equal to done on an object is equal to its its change in kinetic energychange in kinetic energyrdquordquo

22

2

1

2

1if mmW vv

W = KE

The Work-Energy TheoremThe Work-Energy Theorem

bull NoteNotendash The net work done on an object is the work The net work done on an object is the work

done by the net forcedone by the net forcendash Units of energy and work must be the Units of energy and work must be the

same (same (JJ))

W = W = KEKE

II Kinetic energyKinetic energy

IIII WorkWork

IIIIII Work - Kinetic energy theoremWork - Kinetic energy theorem

IVIV Work done by a constant forceWork done by a constant force

- Gravitational force- Gravitational force

V Work done by a variable forceV Work done by a variable force

- Spring force- Spring force

1D-Analysis1D-Analysis

3D-Analysis3D-Analysis

VI PowerVI Power

EnergyEnergy scalar quantity associated with a state scalar quantity associated with a state (or condition) of one or more objects (or condition) of one or more objects

II Kinetic energyKinetic energy

Energy associated with the state of motion of an objectEnergy associated with the state of motion of an object

2

2

1mvK

UnitsUnits 1 Joule = 1J = 1 kgm1 Joule = 1J = 1 kgm22ss22

II WorkII Work

Energy transferred ldquotordquo or ldquofromrdquo an Energy transferred ldquotordquo or ldquofromrdquo an object by means of a force acting on the object by means of a force acting on the object object

To To +W+WFrom From -W-W

- - Constant Constant forceforce

xx maF

dFWdFKKvvm

dvvmmaF

d

vvadavv

xxiff

fxxf

xxf

)(2

1

1)(

2

1

22

20

2

20

220

220

2

Work done by the force = Energy transfer due to the forceWork done by the force = Energy transfer due to the force

-To calculate the work done on an object by a force during To calculate the work done on an object by a force during a displacement we use only the force component alonga displacement we use only the force component along the objectrsquos displacement The force component the objectrsquos displacement The force component perpendicular to the displacement does zero workperpendicular to the displacement does zero work

dFFddFW x

cos

Assumptions Assumptions 1) F = constant force 1) F = constant force 2) Object is particle-like (rigid object all 2) Object is particle-like (rigid object all parts of the object must move together)parts of the object must move together)

090

90180

90

W

W

A force doesA force does +W+W when it has a vector component in the when it has a vector component in the same direction as the displacement andsame direction as the displacement and ndashWndashW when it has when it has a vector component in the opposite directiona vector component in the opposite direction W=0W=0 when it when it has no such vector componenthas no such vector component

Net work done by several forces = Sum of works done by Net work done by several forces = Sum of works done by individual forcesindividual forces CalculationCalculation 1) W1) Wnetnet= W= W11+W+W22+W+W33+hellip+hellip

2) F2) Fnet net W Wnetnet=F=Fnetnet d d

II Work-Kinetic Energy TheoremII Work-Kinetic Energy Theorem

WKKK if

Change in the kinetic energy of the Change in the kinetic energy of the particle particle == Net work done on the particle Net work done on the particle

III Work done by a constant forceIII Work done by a constant force

- - Gravitational forceGravitational force

cosmgdW

Rising objectRising object W= mgd cos180W= mgd cos180ordm ordm = -mgd= -mgd FFgg transfers transfers mgdmgd energyenergy from from the objectrsquos kinetic energy the objectrsquos kinetic energy

Falling down objectFalling down object W= mgd cos 0W= mgd cos 0ordm = +mgdordm = +mgd Fg transfers transfers mgd energyenergy to to the objectrsquos kinetic energy the objectrsquos kinetic energy

External applied force + Gravitational force External applied force + Gravitational force

gaif WWKKK

Object stationary before Object stationary before and after the liftand after the lift Wa+Wg=0

The applied force transfers the same The applied force transfers the same amount of energy to the object as the amount of energy to the object as the gravitational force transfers from the objectgravitational force transfers from the object

IV Work done by a variable forceIV Work done by a variable force

- - Spring forceSpring force

xkF

Hookersquos lawHookersquos law

k = spring constant = spring constant measures springrsquos stiffness measures springrsquos stiffness Units Nm

Hookersquos LawHookersquos Law

When When xx is positive is positive (spring is stretched)(spring is stretched) FF is negativeis negative

WhenWhen xx is is 00 (at the (at the equilibrium position)equilibrium position) F F isis 00

When When xx is negative is negative (spring is compressed)(spring is compressed) FF is positiveis positive

Hookersquos LawHookersquos Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the restoring forceis called the restoring force If the block is released it will If the block is released it will

oscillate back and forth between oscillate back and forth between ndashndashxx and and xx

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

SystemsSystems

A A systemsystem is a small portion of the is a small portion of the UniverseUniversebull We will ignore the details of the rest of the We will ignore the details of the rest of the

UniverseUniverse

A critical skill is to identify the systemA critical skill is to identify the system

Valid SystemValid System

A valid system mayA valid system maybull be a single object or particlebe a single object or particlebull be a collection of objects or particlesbe a collection of objects or particlesbull be a region of spacebe a region of spacebull vary in size and shapevary in size and shape

EnvironmentEnvironment

There is a There is a system boundarysystem boundary around the around the systemsystembull The boundary is an imaginary surfaceThe boundary is an imaginary surfacebull It does not necessarily correspond to a physical It does not necessarily correspond to a physical

boundaryboundary

The boundary divides the system from the The boundary divides the system from the environmentenvironmentbull The environment is the rest of the UniverseThe environment is the rest of the Universe

WorkWork

The workThe work WW done on a system by an done on a system by an agent exerting a constant force on the agent exerting a constant force on the system is the product of the magnitudesystem is the product of the magnitude FF of the of the forceforce the magnitude the magnitude ΔΔrr of the of the displacementdisplacement of the point of application of of the point of application of the force andthe force and cos cos θθ where where θθ is the angle is the angle between the force and the displacement between the force and the displacement vectorsvectors

WorkWork

W = F W = F ΔΔr cos r cos θθbull The displacement is that of the point of The displacement is that of the point of

application of the forceapplication of the forcebull A force does no work on the object if the A force does no work on the object if the

object does not move through a displacementobject does not move through a displacementbull The work done by a force on a moving object The work done by a force on a moving object

is zero when the force applied is is zero when the force applied is perpendicular to the displacement of its point perpendicular to the displacement of its point of applicationof application

Work ExampleWork Example

The normal forceThe normal force nn and and the gravitational forcethe gravitational force mgmg do no work on the objectdo no work on the object

cos cos θθ = cos 90deg= 0= cos 90deg= 0

The forceThe force FF does do work does do work on the objecton the object

More About WorkMore About Work The system and the environment must be The system and the environment must be

determined when dealing with workdetermined when dealing with work The environment does work on the systemThe environment does work on the system

(Work (Work byby the environment the environment onon the system) the system) The sign of the work depends on the direction ofThe sign of the work depends on the direction of

FF relative torelative to ΔΔrr Work is positive when projection ofWork is positive when projection of FF ontoonto ΔΔrr

is in the same direction as the displacementis in the same direction as the displacement Work is negative when the projection is in the Work is negative when the projection is in the

opposite directionopposite direction

Work Is An Energy TransferWork Is An Energy Transfer

bull This is important for a system approach This is important for a system approach to solving a problemto solving a problem

bull If the work is done on a system and it is If the work is done on a system and it is positive energy is transferred to the positive energy is transferred to the systemsystem

bull If the work done on the system is If the work done on the system is negative energy is transferred from the negative energy is transferred from the systemsystem

Work done by a constant forceWork done by a constant force

bull Work done on an object by a constant force Work done on an object by a constant force is defined to be the product of the magnitude of is defined to be the product of the magnitude of the the displacementdisplacement and the component of the and the component of the forceforce parallel to the displacement parallel to the displacement

bull WhereWhere FFIIII is the component of the forceis the component of the force FF parallel to the displacementparallel to the displacement dd

W = FW = FII II middotmiddot d d

WorkIn other words - In other words -

WhereWhere θθ is the angle betweenis the angle between FF and and dd

IfIf θθ is is gt 90gt 90oo work is negative A decelerating car has work is negative A decelerating car has negative work done on it by its enginenegative work done on it by its engine

The unit of work is called The unit of work is called Joule (J)Joule (J) 1J = 1 N1J = 1 Nmiddotmiddotmm

1J=kg∙m1J=kg∙m22ss22

d

F

W = F d cosW = F d cos

Scalar Product of Two VectorsScalar Product of Two Vectors

The scalar product The scalar product of two vectors is of two vectors is written aswritten as A A BB

It is also called It is also called the dot productthe dot product

A A bull bull B = A B cos B = A B cos θθ

θθ is the angle is the angle betweenbetween AA andand BB

Scalar ProductScalar Product

The scalar product is commutativeThe scalar product is commutative A A B = B B = B AA

The scalar product obeys the distributive law The scalar product obeys the distributive law of multiplicationof multiplication

A A (B + C) (B + C) == A A B + A B + A CC

Dot Products of Unit VectorsDot Products of Unit Vectors

bull Using component form withUsing component form with A A andand BB

0kjkiji

1kkjjii

zzyyxx

zyx

zyx

BABABABA

kBjBiBB

kAjAiAA

Work - on and byWork - on and by

A person pushes blockA person pushes block 30 m30 m along the along the ground by exerting force ofground by exerting force of 25 N25 N on the on the trolley How much work does the person do trolley How much work does the person do on the trolleyon the trolley

W = FW = Fd = 25N x 30m = 750 Nmd = 25N x 30m = 750 Nm

Trolley does -Trolley does -750 Nm750 Nm work on the personwork on the person

FFpptFFtrtr

d

A force acts on an object as A force acts on an object as the object moves in the the object moves in the xx direction from the origin direction from the origin to to xx = 500 m = 500 m Find the work Find the work W W done on the done on the object by the forceobject by the force

F 4x ˆ i 3yˆ j N

Mechanical EnergyMechanical Energy

Mechanical energy (energy associated Mechanical energy (energy associated

with with masses) can be thought of as having masses) can be thought of as having

two two components components kinetickinetic and and potentialpotential

ndashKinetic energy is energy of motionKinetic energy is energy of motion

ndashPotential energy is energy of positionPotential energy is energy of position

Kinetic EnergyKinetic Energy

bull In many situations energy can be In many situations energy can be considered asconsidered as ldquothe ability to do workrdquoldquothe ability to do workrdquo

bull Energy can exist in different formsEnergy can exist in different forms

bull Kinetic energy is the energy associated Kinetic energy is the energy associated with the with the motionmotion of an object of an object

bull A moving object can do work on A moving object can do work on another objectanother object

Eg hammer does work on nailEg hammer does work on nail

Kinetic EnergyKinetic Energy Consider an object with massConsider an object with mass mm moving in a straight line with moving in a straight line with initial velocityinitial velocity vvii To accelerate it uniformly to a speedTo accelerate it uniformly to a speed vvff a a constant net forceconstant net force FF is exerted on it parallel to motion over a is exerted on it parallel to motion over a distancedistance dd

Work done on objectWork done on object

W = F d = m a dW = F d = m a d (NII)(NII)SoSo

da

ada

if

iifif

2

2)(222

222

vv

vxxvv

dd

mW if

2

22 vv

Kinetic EnergyKinetic Energy

If we rearrange this we obtainIf we rearrange this we obtain

We define the quantityWe define the quantity frac12mvfrac12mv22 to be the to be the translational translational kinetic energykinetic energy (KE) of the object (KE) of the object

This is the This is the lsquolsquoWork-Energy TheoremWork-Energy Theoremrsquorsquo

ldquoldquoThe net The net workwork done on an object is equal to done on an object is equal to its its change in kinetic energychange in kinetic energyrdquordquo

22

2

1

2

1if mmW vv

W = KE

The Work-Energy TheoremThe Work-Energy Theorem

bull NoteNotendash The net work done on an object is the work The net work done on an object is the work

done by the net forcedone by the net forcendash Units of energy and work must be the Units of energy and work must be the

same (same (JJ))

W = W = KEKE

II Kinetic energyKinetic energy

IIII WorkWork

IIIIII Work - Kinetic energy theoremWork - Kinetic energy theorem

IVIV Work done by a constant forceWork done by a constant force

- Gravitational force- Gravitational force

V Work done by a variable forceV Work done by a variable force

- Spring force- Spring force

1D-Analysis1D-Analysis

3D-Analysis3D-Analysis

VI PowerVI Power

EnergyEnergy scalar quantity associated with a state scalar quantity associated with a state (or condition) of one or more objects (or condition) of one or more objects

II Kinetic energyKinetic energy

Energy associated with the state of motion of an objectEnergy associated with the state of motion of an object

2

2

1mvK

UnitsUnits 1 Joule = 1J = 1 kgm1 Joule = 1J = 1 kgm22ss22

II WorkII Work

Energy transferred ldquotordquo or ldquofromrdquo an Energy transferred ldquotordquo or ldquofromrdquo an object by means of a force acting on the object by means of a force acting on the object object

To To +W+WFrom From -W-W

- - Constant Constant forceforce

xx maF

dFWdFKKvvm

dvvmmaF

d

vvadavv

xxiff

fxxf

xxf

)(2

1

1)(

2

1

22

20

2

20

220

220

2

Work done by the force = Energy transfer due to the forceWork done by the force = Energy transfer due to the force

-To calculate the work done on an object by a force during To calculate the work done on an object by a force during a displacement we use only the force component alonga displacement we use only the force component along the objectrsquos displacement The force component the objectrsquos displacement The force component perpendicular to the displacement does zero workperpendicular to the displacement does zero work

dFFddFW x

cos

Assumptions Assumptions 1) F = constant force 1) F = constant force 2) Object is particle-like (rigid object all 2) Object is particle-like (rigid object all parts of the object must move together)parts of the object must move together)

090

90180

90

W

W

A force doesA force does +W+W when it has a vector component in the when it has a vector component in the same direction as the displacement andsame direction as the displacement and ndashWndashW when it has when it has a vector component in the opposite directiona vector component in the opposite direction W=0W=0 when it when it has no such vector componenthas no such vector component

Net work done by several forces = Sum of works done by Net work done by several forces = Sum of works done by individual forcesindividual forces CalculationCalculation 1) W1) Wnetnet= W= W11+W+W22+W+W33+hellip+hellip

2) F2) Fnet net W Wnetnet=F=Fnetnet d d

II Work-Kinetic Energy TheoremII Work-Kinetic Energy Theorem

WKKK if

Change in the kinetic energy of the Change in the kinetic energy of the particle particle == Net work done on the particle Net work done on the particle

III Work done by a constant forceIII Work done by a constant force

- - Gravitational forceGravitational force

cosmgdW

Rising objectRising object W= mgd cos180W= mgd cos180ordm ordm = -mgd= -mgd FFgg transfers transfers mgdmgd energyenergy from from the objectrsquos kinetic energy the objectrsquos kinetic energy

Falling down objectFalling down object W= mgd cos 0W= mgd cos 0ordm = +mgdordm = +mgd Fg transfers transfers mgd energyenergy to to the objectrsquos kinetic energy the objectrsquos kinetic energy

External applied force + Gravitational force External applied force + Gravitational force

gaif WWKKK

Object stationary before Object stationary before and after the liftand after the lift Wa+Wg=0

The applied force transfers the same The applied force transfers the same amount of energy to the object as the amount of energy to the object as the gravitational force transfers from the objectgravitational force transfers from the object

IV Work done by a variable forceIV Work done by a variable force

- - Spring forceSpring force

xkF

Hookersquos lawHookersquos law

k = spring constant = spring constant measures springrsquos stiffness measures springrsquos stiffness Units Nm

Hookersquos LawHookersquos Law

When When xx is positive is positive (spring is stretched)(spring is stretched) FF is negativeis negative

WhenWhen xx is is 00 (at the (at the equilibrium position)equilibrium position) F F isis 00

When When xx is negative is negative (spring is compressed)(spring is compressed) FF is positiveis positive

Hookersquos LawHookersquos Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the restoring forceis called the restoring force If the block is released it will If the block is released it will

oscillate back and forth between oscillate back and forth between ndashndashxx and and xx

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

Valid SystemValid System

A valid system mayA valid system maybull be a single object or particlebe a single object or particlebull be a collection of objects or particlesbe a collection of objects or particlesbull be a region of spacebe a region of spacebull vary in size and shapevary in size and shape

EnvironmentEnvironment

There is a There is a system boundarysystem boundary around the around the systemsystembull The boundary is an imaginary surfaceThe boundary is an imaginary surfacebull It does not necessarily correspond to a physical It does not necessarily correspond to a physical

boundaryboundary

The boundary divides the system from the The boundary divides the system from the environmentenvironmentbull The environment is the rest of the UniverseThe environment is the rest of the Universe

WorkWork

The workThe work WW done on a system by an done on a system by an agent exerting a constant force on the agent exerting a constant force on the system is the product of the magnitudesystem is the product of the magnitude FF of the of the forceforce the magnitude the magnitude ΔΔrr of the of the displacementdisplacement of the point of application of of the point of application of the force andthe force and cos cos θθ where where θθ is the angle is the angle between the force and the displacement between the force and the displacement vectorsvectors

WorkWork

W = F W = F ΔΔr cos r cos θθbull The displacement is that of the point of The displacement is that of the point of

application of the forceapplication of the forcebull A force does no work on the object if the A force does no work on the object if the

object does not move through a displacementobject does not move through a displacementbull The work done by a force on a moving object The work done by a force on a moving object

is zero when the force applied is is zero when the force applied is perpendicular to the displacement of its point perpendicular to the displacement of its point of applicationof application

Work ExampleWork Example

The normal forceThe normal force nn and and the gravitational forcethe gravitational force mgmg do no work on the objectdo no work on the object

cos cos θθ = cos 90deg= 0= cos 90deg= 0

The forceThe force FF does do work does do work on the objecton the object

More About WorkMore About Work The system and the environment must be The system and the environment must be

determined when dealing with workdetermined when dealing with work The environment does work on the systemThe environment does work on the system

(Work (Work byby the environment the environment onon the system) the system) The sign of the work depends on the direction ofThe sign of the work depends on the direction of

FF relative torelative to ΔΔrr Work is positive when projection ofWork is positive when projection of FF ontoonto ΔΔrr

is in the same direction as the displacementis in the same direction as the displacement Work is negative when the projection is in the Work is negative when the projection is in the

opposite directionopposite direction

Work Is An Energy TransferWork Is An Energy Transfer

bull This is important for a system approach This is important for a system approach to solving a problemto solving a problem

bull If the work is done on a system and it is If the work is done on a system and it is positive energy is transferred to the positive energy is transferred to the systemsystem

bull If the work done on the system is If the work done on the system is negative energy is transferred from the negative energy is transferred from the systemsystem

Work done by a constant forceWork done by a constant force

bull Work done on an object by a constant force Work done on an object by a constant force is defined to be the product of the magnitude of is defined to be the product of the magnitude of the the displacementdisplacement and the component of the and the component of the forceforce parallel to the displacement parallel to the displacement

bull WhereWhere FFIIII is the component of the forceis the component of the force FF parallel to the displacementparallel to the displacement dd

W = FW = FII II middotmiddot d d

WorkIn other words - In other words -

WhereWhere θθ is the angle betweenis the angle between FF and and dd

IfIf θθ is is gt 90gt 90oo work is negative A decelerating car has work is negative A decelerating car has negative work done on it by its enginenegative work done on it by its engine

The unit of work is called The unit of work is called Joule (J)Joule (J) 1J = 1 N1J = 1 Nmiddotmiddotmm

1J=kg∙m1J=kg∙m22ss22

d

F

W = F d cosW = F d cos

Scalar Product of Two VectorsScalar Product of Two Vectors

The scalar product The scalar product of two vectors is of two vectors is written aswritten as A A BB

It is also called It is also called the dot productthe dot product

A A bull bull B = A B cos B = A B cos θθ

θθ is the angle is the angle betweenbetween AA andand BB

Scalar ProductScalar Product

The scalar product is commutativeThe scalar product is commutative A A B = B B = B AA

The scalar product obeys the distributive law The scalar product obeys the distributive law of multiplicationof multiplication

A A (B + C) (B + C) == A A B + A B + A CC

Dot Products of Unit VectorsDot Products of Unit Vectors

bull Using component form withUsing component form with A A andand BB

0kjkiji

1kkjjii

zzyyxx

zyx

zyx

BABABABA

kBjBiBB

kAjAiAA

Work - on and byWork - on and by

A person pushes blockA person pushes block 30 m30 m along the along the ground by exerting force ofground by exerting force of 25 N25 N on the on the trolley How much work does the person do trolley How much work does the person do on the trolleyon the trolley

W = FW = Fd = 25N x 30m = 750 Nmd = 25N x 30m = 750 Nm

Trolley does -Trolley does -750 Nm750 Nm work on the personwork on the person

FFpptFFtrtr

d

A force acts on an object as A force acts on an object as the object moves in the the object moves in the xx direction from the origin direction from the origin to to xx = 500 m = 500 m Find the work Find the work W W done on the done on the object by the forceobject by the force

F 4x ˆ i 3yˆ j N

Mechanical EnergyMechanical Energy

Mechanical energy (energy associated Mechanical energy (energy associated

with with masses) can be thought of as having masses) can be thought of as having

two two components components kinetickinetic and and potentialpotential

ndashKinetic energy is energy of motionKinetic energy is energy of motion

ndashPotential energy is energy of positionPotential energy is energy of position

Kinetic EnergyKinetic Energy

bull In many situations energy can be In many situations energy can be considered asconsidered as ldquothe ability to do workrdquoldquothe ability to do workrdquo

bull Energy can exist in different formsEnergy can exist in different forms

bull Kinetic energy is the energy associated Kinetic energy is the energy associated with the with the motionmotion of an object of an object

bull A moving object can do work on A moving object can do work on another objectanother object

Eg hammer does work on nailEg hammer does work on nail

Kinetic EnergyKinetic Energy Consider an object with massConsider an object with mass mm moving in a straight line with moving in a straight line with initial velocityinitial velocity vvii To accelerate it uniformly to a speedTo accelerate it uniformly to a speed vvff a a constant net forceconstant net force FF is exerted on it parallel to motion over a is exerted on it parallel to motion over a distancedistance dd

Work done on objectWork done on object

W = F d = m a dW = F d = m a d (NII)(NII)SoSo

da

ada

if

iifif

2

2)(222

222

vv

vxxvv

dd

mW if

2

22 vv

Kinetic EnergyKinetic Energy

If we rearrange this we obtainIf we rearrange this we obtain

We define the quantityWe define the quantity frac12mvfrac12mv22 to be the to be the translational translational kinetic energykinetic energy (KE) of the object (KE) of the object

This is the This is the lsquolsquoWork-Energy TheoremWork-Energy Theoremrsquorsquo

ldquoldquoThe net The net workwork done on an object is equal to done on an object is equal to its its change in kinetic energychange in kinetic energyrdquordquo

22

2

1

2

1if mmW vv

W = KE

The Work-Energy TheoremThe Work-Energy Theorem

bull NoteNotendash The net work done on an object is the work The net work done on an object is the work

done by the net forcedone by the net forcendash Units of energy and work must be the Units of energy and work must be the

same (same (JJ))

W = W = KEKE

II Kinetic energyKinetic energy

IIII WorkWork

IIIIII Work - Kinetic energy theoremWork - Kinetic energy theorem

IVIV Work done by a constant forceWork done by a constant force

- Gravitational force- Gravitational force

V Work done by a variable forceV Work done by a variable force

- Spring force- Spring force

1D-Analysis1D-Analysis

3D-Analysis3D-Analysis

VI PowerVI Power

EnergyEnergy scalar quantity associated with a state scalar quantity associated with a state (or condition) of one or more objects (or condition) of one or more objects

II Kinetic energyKinetic energy

Energy associated with the state of motion of an objectEnergy associated with the state of motion of an object

2

2

1mvK

UnitsUnits 1 Joule = 1J = 1 kgm1 Joule = 1J = 1 kgm22ss22

II WorkII Work

Energy transferred ldquotordquo or ldquofromrdquo an Energy transferred ldquotordquo or ldquofromrdquo an object by means of a force acting on the object by means of a force acting on the object object

To To +W+WFrom From -W-W

- - Constant Constant forceforce

xx maF

dFWdFKKvvm

dvvmmaF

d

vvadavv

xxiff

fxxf

xxf

)(2

1

1)(

2

1

22

20

2

20

220

220

2

Work done by the force = Energy transfer due to the forceWork done by the force = Energy transfer due to the force

-To calculate the work done on an object by a force during To calculate the work done on an object by a force during a displacement we use only the force component alonga displacement we use only the force component along the objectrsquos displacement The force component the objectrsquos displacement The force component perpendicular to the displacement does zero workperpendicular to the displacement does zero work

dFFddFW x

cos

Assumptions Assumptions 1) F = constant force 1) F = constant force 2) Object is particle-like (rigid object all 2) Object is particle-like (rigid object all parts of the object must move together)parts of the object must move together)

090

90180

90

W

W

A force doesA force does +W+W when it has a vector component in the when it has a vector component in the same direction as the displacement andsame direction as the displacement and ndashWndashW when it has when it has a vector component in the opposite directiona vector component in the opposite direction W=0W=0 when it when it has no such vector componenthas no such vector component

Net work done by several forces = Sum of works done by Net work done by several forces = Sum of works done by individual forcesindividual forces CalculationCalculation 1) W1) Wnetnet= W= W11+W+W22+W+W33+hellip+hellip

2) F2) Fnet net W Wnetnet=F=Fnetnet d d

II Work-Kinetic Energy TheoremII Work-Kinetic Energy Theorem

WKKK if

Change in the kinetic energy of the Change in the kinetic energy of the particle particle == Net work done on the particle Net work done on the particle

III Work done by a constant forceIII Work done by a constant force

- - Gravitational forceGravitational force

cosmgdW

Rising objectRising object W= mgd cos180W= mgd cos180ordm ordm = -mgd= -mgd FFgg transfers transfers mgdmgd energyenergy from from the objectrsquos kinetic energy the objectrsquos kinetic energy

Falling down objectFalling down object W= mgd cos 0W= mgd cos 0ordm = +mgdordm = +mgd Fg transfers transfers mgd energyenergy to to the objectrsquos kinetic energy the objectrsquos kinetic energy

External applied force + Gravitational force External applied force + Gravitational force

gaif WWKKK

Object stationary before Object stationary before and after the liftand after the lift Wa+Wg=0

The applied force transfers the same The applied force transfers the same amount of energy to the object as the amount of energy to the object as the gravitational force transfers from the objectgravitational force transfers from the object

IV Work done by a variable forceIV Work done by a variable force

- - Spring forceSpring force

xkF

Hookersquos lawHookersquos law

k = spring constant = spring constant measures springrsquos stiffness measures springrsquos stiffness Units Nm

Hookersquos LawHookersquos Law

When When xx is positive is positive (spring is stretched)(spring is stretched) FF is negativeis negative

WhenWhen xx is is 00 (at the (at the equilibrium position)equilibrium position) F F isis 00

When When xx is negative is negative (spring is compressed)(spring is compressed) FF is positiveis positive

Hookersquos LawHookersquos Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the restoring forceis called the restoring force If the block is released it will If the block is released it will

oscillate back and forth between oscillate back and forth between ndashndashxx and and xx

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

EnvironmentEnvironment

There is a There is a system boundarysystem boundary around the around the systemsystembull The boundary is an imaginary surfaceThe boundary is an imaginary surfacebull It does not necessarily correspond to a physical It does not necessarily correspond to a physical

boundaryboundary

The boundary divides the system from the The boundary divides the system from the environmentenvironmentbull The environment is the rest of the UniverseThe environment is the rest of the Universe

WorkWork

The workThe work WW done on a system by an done on a system by an agent exerting a constant force on the agent exerting a constant force on the system is the product of the magnitudesystem is the product of the magnitude FF of the of the forceforce the magnitude the magnitude ΔΔrr of the of the displacementdisplacement of the point of application of of the point of application of the force andthe force and cos cos θθ where where θθ is the angle is the angle between the force and the displacement between the force and the displacement vectorsvectors

WorkWork

W = F W = F ΔΔr cos r cos θθbull The displacement is that of the point of The displacement is that of the point of

application of the forceapplication of the forcebull A force does no work on the object if the A force does no work on the object if the

object does not move through a displacementobject does not move through a displacementbull The work done by a force on a moving object The work done by a force on a moving object

is zero when the force applied is is zero when the force applied is perpendicular to the displacement of its point perpendicular to the displacement of its point of applicationof application

Work ExampleWork Example

The normal forceThe normal force nn and and the gravitational forcethe gravitational force mgmg do no work on the objectdo no work on the object

cos cos θθ = cos 90deg= 0= cos 90deg= 0

The forceThe force FF does do work does do work on the objecton the object

More About WorkMore About Work The system and the environment must be The system and the environment must be

determined when dealing with workdetermined when dealing with work The environment does work on the systemThe environment does work on the system

(Work (Work byby the environment the environment onon the system) the system) The sign of the work depends on the direction ofThe sign of the work depends on the direction of

FF relative torelative to ΔΔrr Work is positive when projection ofWork is positive when projection of FF ontoonto ΔΔrr

is in the same direction as the displacementis in the same direction as the displacement Work is negative when the projection is in the Work is negative when the projection is in the

opposite directionopposite direction

Work Is An Energy TransferWork Is An Energy Transfer

bull This is important for a system approach This is important for a system approach to solving a problemto solving a problem

bull If the work is done on a system and it is If the work is done on a system and it is positive energy is transferred to the positive energy is transferred to the systemsystem

bull If the work done on the system is If the work done on the system is negative energy is transferred from the negative energy is transferred from the systemsystem

Work done by a constant forceWork done by a constant force

bull Work done on an object by a constant force Work done on an object by a constant force is defined to be the product of the magnitude of is defined to be the product of the magnitude of the the displacementdisplacement and the component of the and the component of the forceforce parallel to the displacement parallel to the displacement

bull WhereWhere FFIIII is the component of the forceis the component of the force FF parallel to the displacementparallel to the displacement dd

W = FW = FII II middotmiddot d d

WorkIn other words - In other words -

WhereWhere θθ is the angle betweenis the angle between FF and and dd

IfIf θθ is is gt 90gt 90oo work is negative A decelerating car has work is negative A decelerating car has negative work done on it by its enginenegative work done on it by its engine

The unit of work is called The unit of work is called Joule (J)Joule (J) 1J = 1 N1J = 1 Nmiddotmiddotmm

1J=kg∙m1J=kg∙m22ss22

d

F

W = F d cosW = F d cos

Scalar Product of Two VectorsScalar Product of Two Vectors

The scalar product The scalar product of two vectors is of two vectors is written aswritten as A A BB

It is also called It is also called the dot productthe dot product

A A bull bull B = A B cos B = A B cos θθ

θθ is the angle is the angle betweenbetween AA andand BB

Scalar ProductScalar Product

The scalar product is commutativeThe scalar product is commutative A A B = B B = B AA

The scalar product obeys the distributive law The scalar product obeys the distributive law of multiplicationof multiplication

A A (B + C) (B + C) == A A B + A B + A CC

Dot Products of Unit VectorsDot Products of Unit Vectors

bull Using component form withUsing component form with A A andand BB

0kjkiji

1kkjjii

zzyyxx

zyx

zyx

BABABABA

kBjBiBB

kAjAiAA

Work - on and byWork - on and by

A person pushes blockA person pushes block 30 m30 m along the along the ground by exerting force ofground by exerting force of 25 N25 N on the on the trolley How much work does the person do trolley How much work does the person do on the trolleyon the trolley

W = FW = Fd = 25N x 30m = 750 Nmd = 25N x 30m = 750 Nm

Trolley does -Trolley does -750 Nm750 Nm work on the personwork on the person

FFpptFFtrtr

d

A force acts on an object as A force acts on an object as the object moves in the the object moves in the xx direction from the origin direction from the origin to to xx = 500 m = 500 m Find the work Find the work W W done on the done on the object by the forceobject by the force

F 4x ˆ i 3yˆ j N

Mechanical EnergyMechanical Energy

Mechanical energy (energy associated Mechanical energy (energy associated

with with masses) can be thought of as having masses) can be thought of as having

two two components components kinetickinetic and and potentialpotential

ndashKinetic energy is energy of motionKinetic energy is energy of motion

ndashPotential energy is energy of positionPotential energy is energy of position

Kinetic EnergyKinetic Energy

bull In many situations energy can be In many situations energy can be considered asconsidered as ldquothe ability to do workrdquoldquothe ability to do workrdquo

bull Energy can exist in different formsEnergy can exist in different forms

bull Kinetic energy is the energy associated Kinetic energy is the energy associated with the with the motionmotion of an object of an object

bull A moving object can do work on A moving object can do work on another objectanother object

Eg hammer does work on nailEg hammer does work on nail

Kinetic EnergyKinetic Energy Consider an object with massConsider an object with mass mm moving in a straight line with moving in a straight line with initial velocityinitial velocity vvii To accelerate it uniformly to a speedTo accelerate it uniformly to a speed vvff a a constant net forceconstant net force FF is exerted on it parallel to motion over a is exerted on it parallel to motion over a distancedistance dd

Work done on objectWork done on object

W = F d = m a dW = F d = m a d (NII)(NII)SoSo

da

ada

if

iifif

2

2)(222

222

vv

vxxvv

dd

mW if

2

22 vv

Kinetic EnergyKinetic Energy

If we rearrange this we obtainIf we rearrange this we obtain

We define the quantityWe define the quantity frac12mvfrac12mv22 to be the to be the translational translational kinetic energykinetic energy (KE) of the object (KE) of the object

This is the This is the lsquolsquoWork-Energy TheoremWork-Energy Theoremrsquorsquo

ldquoldquoThe net The net workwork done on an object is equal to done on an object is equal to its its change in kinetic energychange in kinetic energyrdquordquo

22

2

1

2

1if mmW vv

W = KE

The Work-Energy TheoremThe Work-Energy Theorem

bull NoteNotendash The net work done on an object is the work The net work done on an object is the work

done by the net forcedone by the net forcendash Units of energy and work must be the Units of energy and work must be the

same (same (JJ))

W = W = KEKE

II Kinetic energyKinetic energy

IIII WorkWork

IIIIII Work - Kinetic energy theoremWork - Kinetic energy theorem

IVIV Work done by a constant forceWork done by a constant force

- Gravitational force- Gravitational force

V Work done by a variable forceV Work done by a variable force

- Spring force- Spring force

1D-Analysis1D-Analysis

3D-Analysis3D-Analysis

VI PowerVI Power

EnergyEnergy scalar quantity associated with a state scalar quantity associated with a state (or condition) of one or more objects (or condition) of one or more objects

II Kinetic energyKinetic energy

Energy associated with the state of motion of an objectEnergy associated with the state of motion of an object

2

2

1mvK

UnitsUnits 1 Joule = 1J = 1 kgm1 Joule = 1J = 1 kgm22ss22

II WorkII Work

Energy transferred ldquotordquo or ldquofromrdquo an Energy transferred ldquotordquo or ldquofromrdquo an object by means of a force acting on the object by means of a force acting on the object object

To To +W+WFrom From -W-W

- - Constant Constant forceforce

xx maF

dFWdFKKvvm

dvvmmaF

d

vvadavv

xxiff

fxxf

xxf

)(2

1

1)(

2

1

22

20

2

20

220

220

2

Work done by the force = Energy transfer due to the forceWork done by the force = Energy transfer due to the force

-To calculate the work done on an object by a force during To calculate the work done on an object by a force during a displacement we use only the force component alonga displacement we use only the force component along the objectrsquos displacement The force component the objectrsquos displacement The force component perpendicular to the displacement does zero workperpendicular to the displacement does zero work

dFFddFW x

cos

Assumptions Assumptions 1) F = constant force 1) F = constant force 2) Object is particle-like (rigid object all 2) Object is particle-like (rigid object all parts of the object must move together)parts of the object must move together)

090

90180

90

W

W

A force doesA force does +W+W when it has a vector component in the when it has a vector component in the same direction as the displacement andsame direction as the displacement and ndashWndashW when it has when it has a vector component in the opposite directiona vector component in the opposite direction W=0W=0 when it when it has no such vector componenthas no such vector component

Net work done by several forces = Sum of works done by Net work done by several forces = Sum of works done by individual forcesindividual forces CalculationCalculation 1) W1) Wnetnet= W= W11+W+W22+W+W33+hellip+hellip

2) F2) Fnet net W Wnetnet=F=Fnetnet d d

II Work-Kinetic Energy TheoremII Work-Kinetic Energy Theorem

WKKK if

Change in the kinetic energy of the Change in the kinetic energy of the particle particle == Net work done on the particle Net work done on the particle

III Work done by a constant forceIII Work done by a constant force

- - Gravitational forceGravitational force

cosmgdW

Rising objectRising object W= mgd cos180W= mgd cos180ordm ordm = -mgd= -mgd FFgg transfers transfers mgdmgd energyenergy from from the objectrsquos kinetic energy the objectrsquos kinetic energy

Falling down objectFalling down object W= mgd cos 0W= mgd cos 0ordm = +mgdordm = +mgd Fg transfers transfers mgd energyenergy to to the objectrsquos kinetic energy the objectrsquos kinetic energy

External applied force + Gravitational force External applied force + Gravitational force

gaif WWKKK

Object stationary before Object stationary before and after the liftand after the lift Wa+Wg=0

The applied force transfers the same The applied force transfers the same amount of energy to the object as the amount of energy to the object as the gravitational force transfers from the objectgravitational force transfers from the object

IV Work done by a variable forceIV Work done by a variable force

- - Spring forceSpring force

xkF

Hookersquos lawHookersquos law

k = spring constant = spring constant measures springrsquos stiffness measures springrsquos stiffness Units Nm

Hookersquos LawHookersquos Law

When When xx is positive is positive (spring is stretched)(spring is stretched) FF is negativeis negative

WhenWhen xx is is 00 (at the (at the equilibrium position)equilibrium position) F F isis 00

When When xx is negative is negative (spring is compressed)(spring is compressed) FF is positiveis positive

Hookersquos LawHookersquos Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the restoring forceis called the restoring force If the block is released it will If the block is released it will

oscillate back and forth between oscillate back and forth between ndashndashxx and and xx

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

WorkWork

The workThe work WW done on a system by an done on a system by an agent exerting a constant force on the agent exerting a constant force on the system is the product of the magnitudesystem is the product of the magnitude FF of the of the forceforce the magnitude the magnitude ΔΔrr of the of the displacementdisplacement of the point of application of of the point of application of the force andthe force and cos cos θθ where where θθ is the angle is the angle between the force and the displacement between the force and the displacement vectorsvectors

WorkWork

W = F W = F ΔΔr cos r cos θθbull The displacement is that of the point of The displacement is that of the point of

application of the forceapplication of the forcebull A force does no work on the object if the A force does no work on the object if the

object does not move through a displacementobject does not move through a displacementbull The work done by a force on a moving object The work done by a force on a moving object

is zero when the force applied is is zero when the force applied is perpendicular to the displacement of its point perpendicular to the displacement of its point of applicationof application

Work ExampleWork Example

The normal forceThe normal force nn and and the gravitational forcethe gravitational force mgmg do no work on the objectdo no work on the object

cos cos θθ = cos 90deg= 0= cos 90deg= 0

The forceThe force FF does do work does do work on the objecton the object

More About WorkMore About Work The system and the environment must be The system and the environment must be

determined when dealing with workdetermined when dealing with work The environment does work on the systemThe environment does work on the system

(Work (Work byby the environment the environment onon the system) the system) The sign of the work depends on the direction ofThe sign of the work depends on the direction of

FF relative torelative to ΔΔrr Work is positive when projection ofWork is positive when projection of FF ontoonto ΔΔrr

is in the same direction as the displacementis in the same direction as the displacement Work is negative when the projection is in the Work is negative when the projection is in the

opposite directionopposite direction

Work Is An Energy TransferWork Is An Energy Transfer

bull This is important for a system approach This is important for a system approach to solving a problemto solving a problem

bull If the work is done on a system and it is If the work is done on a system and it is positive energy is transferred to the positive energy is transferred to the systemsystem

bull If the work done on the system is If the work done on the system is negative energy is transferred from the negative energy is transferred from the systemsystem

Work done by a constant forceWork done by a constant force

bull Work done on an object by a constant force Work done on an object by a constant force is defined to be the product of the magnitude of is defined to be the product of the magnitude of the the displacementdisplacement and the component of the and the component of the forceforce parallel to the displacement parallel to the displacement

bull WhereWhere FFIIII is the component of the forceis the component of the force FF parallel to the displacementparallel to the displacement dd

W = FW = FII II middotmiddot d d

WorkIn other words - In other words -

WhereWhere θθ is the angle betweenis the angle between FF and and dd

IfIf θθ is is gt 90gt 90oo work is negative A decelerating car has work is negative A decelerating car has negative work done on it by its enginenegative work done on it by its engine

The unit of work is called The unit of work is called Joule (J)Joule (J) 1J = 1 N1J = 1 Nmiddotmiddotmm

1J=kg∙m1J=kg∙m22ss22

d

F

W = F d cosW = F d cos

Scalar Product of Two VectorsScalar Product of Two Vectors

The scalar product The scalar product of two vectors is of two vectors is written aswritten as A A BB

It is also called It is also called the dot productthe dot product

A A bull bull B = A B cos B = A B cos θθ

θθ is the angle is the angle betweenbetween AA andand BB

Scalar ProductScalar Product

The scalar product is commutativeThe scalar product is commutative A A B = B B = B AA

The scalar product obeys the distributive law The scalar product obeys the distributive law of multiplicationof multiplication

A A (B + C) (B + C) == A A B + A B + A CC

Dot Products of Unit VectorsDot Products of Unit Vectors

bull Using component form withUsing component form with A A andand BB

0kjkiji

1kkjjii

zzyyxx

zyx

zyx

BABABABA

kBjBiBB

kAjAiAA

Work - on and byWork - on and by

A person pushes blockA person pushes block 30 m30 m along the along the ground by exerting force ofground by exerting force of 25 N25 N on the on the trolley How much work does the person do trolley How much work does the person do on the trolleyon the trolley

W = FW = Fd = 25N x 30m = 750 Nmd = 25N x 30m = 750 Nm

Trolley does -Trolley does -750 Nm750 Nm work on the personwork on the person

FFpptFFtrtr

d

A force acts on an object as A force acts on an object as the object moves in the the object moves in the xx direction from the origin direction from the origin to to xx = 500 m = 500 m Find the work Find the work W W done on the done on the object by the forceobject by the force

F 4x ˆ i 3yˆ j N

Mechanical EnergyMechanical Energy

Mechanical energy (energy associated Mechanical energy (energy associated

with with masses) can be thought of as having masses) can be thought of as having

two two components components kinetickinetic and and potentialpotential

ndashKinetic energy is energy of motionKinetic energy is energy of motion

ndashPotential energy is energy of positionPotential energy is energy of position

Kinetic EnergyKinetic Energy

bull In many situations energy can be In many situations energy can be considered asconsidered as ldquothe ability to do workrdquoldquothe ability to do workrdquo

bull Energy can exist in different formsEnergy can exist in different forms

bull Kinetic energy is the energy associated Kinetic energy is the energy associated with the with the motionmotion of an object of an object

bull A moving object can do work on A moving object can do work on another objectanother object

Eg hammer does work on nailEg hammer does work on nail

Kinetic EnergyKinetic Energy Consider an object with massConsider an object with mass mm moving in a straight line with moving in a straight line with initial velocityinitial velocity vvii To accelerate it uniformly to a speedTo accelerate it uniformly to a speed vvff a a constant net forceconstant net force FF is exerted on it parallel to motion over a is exerted on it parallel to motion over a distancedistance dd

Work done on objectWork done on object

W = F d = m a dW = F d = m a d (NII)(NII)SoSo

da

ada

if

iifif

2

2)(222

222

vv

vxxvv

dd

mW if

2

22 vv

Kinetic EnergyKinetic Energy

If we rearrange this we obtainIf we rearrange this we obtain

We define the quantityWe define the quantity frac12mvfrac12mv22 to be the to be the translational translational kinetic energykinetic energy (KE) of the object (KE) of the object

This is the This is the lsquolsquoWork-Energy TheoremWork-Energy Theoremrsquorsquo

ldquoldquoThe net The net workwork done on an object is equal to done on an object is equal to its its change in kinetic energychange in kinetic energyrdquordquo

22

2

1

2

1if mmW vv

W = KE

The Work-Energy TheoremThe Work-Energy Theorem

bull NoteNotendash The net work done on an object is the work The net work done on an object is the work

done by the net forcedone by the net forcendash Units of energy and work must be the Units of energy and work must be the

same (same (JJ))

W = W = KEKE

II Kinetic energyKinetic energy

IIII WorkWork

IIIIII Work - Kinetic energy theoremWork - Kinetic energy theorem

IVIV Work done by a constant forceWork done by a constant force

- Gravitational force- Gravitational force

V Work done by a variable forceV Work done by a variable force

- Spring force- Spring force

1D-Analysis1D-Analysis

3D-Analysis3D-Analysis

VI PowerVI Power

EnergyEnergy scalar quantity associated with a state scalar quantity associated with a state (or condition) of one or more objects (or condition) of one or more objects

II Kinetic energyKinetic energy

Energy associated with the state of motion of an objectEnergy associated with the state of motion of an object

2

2

1mvK

UnitsUnits 1 Joule = 1J = 1 kgm1 Joule = 1J = 1 kgm22ss22

II WorkII Work

Energy transferred ldquotordquo or ldquofromrdquo an Energy transferred ldquotordquo or ldquofromrdquo an object by means of a force acting on the object by means of a force acting on the object object

To To +W+WFrom From -W-W

- - Constant Constant forceforce

xx maF

dFWdFKKvvm

dvvmmaF

d

vvadavv

xxiff

fxxf

xxf

)(2

1

1)(

2

1

22

20

2

20

220

220

2

Work done by the force = Energy transfer due to the forceWork done by the force = Energy transfer due to the force

-To calculate the work done on an object by a force during To calculate the work done on an object by a force during a displacement we use only the force component alonga displacement we use only the force component along the objectrsquos displacement The force component the objectrsquos displacement The force component perpendicular to the displacement does zero workperpendicular to the displacement does zero work

dFFddFW x

cos

Assumptions Assumptions 1) F = constant force 1) F = constant force 2) Object is particle-like (rigid object all 2) Object is particle-like (rigid object all parts of the object must move together)parts of the object must move together)

090

90180

90

W

W

A force doesA force does +W+W when it has a vector component in the when it has a vector component in the same direction as the displacement andsame direction as the displacement and ndashWndashW when it has when it has a vector component in the opposite directiona vector component in the opposite direction W=0W=0 when it when it has no such vector componenthas no such vector component

Net work done by several forces = Sum of works done by Net work done by several forces = Sum of works done by individual forcesindividual forces CalculationCalculation 1) W1) Wnetnet= W= W11+W+W22+W+W33+hellip+hellip

2) F2) Fnet net W Wnetnet=F=Fnetnet d d

II Work-Kinetic Energy TheoremII Work-Kinetic Energy Theorem

WKKK if

Change in the kinetic energy of the Change in the kinetic energy of the particle particle == Net work done on the particle Net work done on the particle

III Work done by a constant forceIII Work done by a constant force

- - Gravitational forceGravitational force

cosmgdW

Rising objectRising object W= mgd cos180W= mgd cos180ordm ordm = -mgd= -mgd FFgg transfers transfers mgdmgd energyenergy from from the objectrsquos kinetic energy the objectrsquos kinetic energy

Falling down objectFalling down object W= mgd cos 0W= mgd cos 0ordm = +mgdordm = +mgd Fg transfers transfers mgd energyenergy to to the objectrsquos kinetic energy the objectrsquos kinetic energy

External applied force + Gravitational force External applied force + Gravitational force

gaif WWKKK

Object stationary before Object stationary before and after the liftand after the lift Wa+Wg=0

The applied force transfers the same The applied force transfers the same amount of energy to the object as the amount of energy to the object as the gravitational force transfers from the objectgravitational force transfers from the object

IV Work done by a variable forceIV Work done by a variable force

- - Spring forceSpring force

xkF

Hookersquos lawHookersquos law

k = spring constant = spring constant measures springrsquos stiffness measures springrsquos stiffness Units Nm

Hookersquos LawHookersquos Law

When When xx is positive is positive (spring is stretched)(spring is stretched) FF is negativeis negative

WhenWhen xx is is 00 (at the (at the equilibrium position)equilibrium position) F F isis 00

When When xx is negative is negative (spring is compressed)(spring is compressed) FF is positiveis positive

Hookersquos LawHookersquos Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the restoring forceis called the restoring force If the block is released it will If the block is released it will

oscillate back and forth between oscillate back and forth between ndashndashxx and and xx

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

WorkWork

W = F W = F ΔΔr cos r cos θθbull The displacement is that of the point of The displacement is that of the point of

application of the forceapplication of the forcebull A force does no work on the object if the A force does no work on the object if the

object does not move through a displacementobject does not move through a displacementbull The work done by a force on a moving object The work done by a force on a moving object

is zero when the force applied is is zero when the force applied is perpendicular to the displacement of its point perpendicular to the displacement of its point of applicationof application

Work ExampleWork Example

The normal forceThe normal force nn and and the gravitational forcethe gravitational force mgmg do no work on the objectdo no work on the object

cos cos θθ = cos 90deg= 0= cos 90deg= 0

The forceThe force FF does do work does do work on the objecton the object

More About WorkMore About Work The system and the environment must be The system and the environment must be

determined when dealing with workdetermined when dealing with work The environment does work on the systemThe environment does work on the system

(Work (Work byby the environment the environment onon the system) the system) The sign of the work depends on the direction ofThe sign of the work depends on the direction of

FF relative torelative to ΔΔrr Work is positive when projection ofWork is positive when projection of FF ontoonto ΔΔrr

is in the same direction as the displacementis in the same direction as the displacement Work is negative when the projection is in the Work is negative when the projection is in the

opposite directionopposite direction

Work Is An Energy TransferWork Is An Energy Transfer

bull This is important for a system approach This is important for a system approach to solving a problemto solving a problem

bull If the work is done on a system and it is If the work is done on a system and it is positive energy is transferred to the positive energy is transferred to the systemsystem

bull If the work done on the system is If the work done on the system is negative energy is transferred from the negative energy is transferred from the systemsystem

Work done by a constant forceWork done by a constant force

bull Work done on an object by a constant force Work done on an object by a constant force is defined to be the product of the magnitude of is defined to be the product of the magnitude of the the displacementdisplacement and the component of the and the component of the forceforce parallel to the displacement parallel to the displacement

bull WhereWhere FFIIII is the component of the forceis the component of the force FF parallel to the displacementparallel to the displacement dd

W = FW = FII II middotmiddot d d

WorkIn other words - In other words -

WhereWhere θθ is the angle betweenis the angle between FF and and dd

IfIf θθ is is gt 90gt 90oo work is negative A decelerating car has work is negative A decelerating car has negative work done on it by its enginenegative work done on it by its engine

The unit of work is called The unit of work is called Joule (J)Joule (J) 1J = 1 N1J = 1 Nmiddotmiddotmm

1J=kg∙m1J=kg∙m22ss22

d

F

W = F d cosW = F d cos

Scalar Product of Two VectorsScalar Product of Two Vectors

The scalar product The scalar product of two vectors is of two vectors is written aswritten as A A BB

It is also called It is also called the dot productthe dot product

A A bull bull B = A B cos B = A B cos θθ

θθ is the angle is the angle betweenbetween AA andand BB

Scalar ProductScalar Product

The scalar product is commutativeThe scalar product is commutative A A B = B B = B AA

The scalar product obeys the distributive law The scalar product obeys the distributive law of multiplicationof multiplication

A A (B + C) (B + C) == A A B + A B + A CC

Dot Products of Unit VectorsDot Products of Unit Vectors

bull Using component form withUsing component form with A A andand BB

0kjkiji

1kkjjii

zzyyxx

zyx

zyx

BABABABA

kBjBiBB

kAjAiAA

Work - on and byWork - on and by

A person pushes blockA person pushes block 30 m30 m along the along the ground by exerting force ofground by exerting force of 25 N25 N on the on the trolley How much work does the person do trolley How much work does the person do on the trolleyon the trolley

W = FW = Fd = 25N x 30m = 750 Nmd = 25N x 30m = 750 Nm

Trolley does -Trolley does -750 Nm750 Nm work on the personwork on the person

FFpptFFtrtr

d

A force acts on an object as A force acts on an object as the object moves in the the object moves in the xx direction from the origin direction from the origin to to xx = 500 m = 500 m Find the work Find the work W W done on the done on the object by the forceobject by the force

F 4x ˆ i 3yˆ j N

Mechanical EnergyMechanical Energy

Mechanical energy (energy associated Mechanical energy (energy associated

with with masses) can be thought of as having masses) can be thought of as having

two two components components kinetickinetic and and potentialpotential

ndashKinetic energy is energy of motionKinetic energy is energy of motion

ndashPotential energy is energy of positionPotential energy is energy of position

Kinetic EnergyKinetic Energy

bull In many situations energy can be In many situations energy can be considered asconsidered as ldquothe ability to do workrdquoldquothe ability to do workrdquo

bull Energy can exist in different formsEnergy can exist in different forms

bull Kinetic energy is the energy associated Kinetic energy is the energy associated with the with the motionmotion of an object of an object

bull A moving object can do work on A moving object can do work on another objectanother object

Eg hammer does work on nailEg hammer does work on nail

Kinetic EnergyKinetic Energy Consider an object with massConsider an object with mass mm moving in a straight line with moving in a straight line with initial velocityinitial velocity vvii To accelerate it uniformly to a speedTo accelerate it uniformly to a speed vvff a a constant net forceconstant net force FF is exerted on it parallel to motion over a is exerted on it parallel to motion over a distancedistance dd

Work done on objectWork done on object

W = F d = m a dW = F d = m a d (NII)(NII)SoSo

da

ada

if

iifif

2

2)(222

222

vv

vxxvv

dd

mW if

2

22 vv

Kinetic EnergyKinetic Energy

If we rearrange this we obtainIf we rearrange this we obtain

We define the quantityWe define the quantity frac12mvfrac12mv22 to be the to be the translational translational kinetic energykinetic energy (KE) of the object (KE) of the object

This is the This is the lsquolsquoWork-Energy TheoremWork-Energy Theoremrsquorsquo

ldquoldquoThe net The net workwork done on an object is equal to done on an object is equal to its its change in kinetic energychange in kinetic energyrdquordquo

22

2

1

2

1if mmW vv

W = KE

The Work-Energy TheoremThe Work-Energy Theorem

bull NoteNotendash The net work done on an object is the work The net work done on an object is the work

done by the net forcedone by the net forcendash Units of energy and work must be the Units of energy and work must be the

same (same (JJ))

W = W = KEKE

II Kinetic energyKinetic energy

IIII WorkWork

IIIIII Work - Kinetic energy theoremWork - Kinetic energy theorem

IVIV Work done by a constant forceWork done by a constant force

- Gravitational force- Gravitational force

V Work done by a variable forceV Work done by a variable force

- Spring force- Spring force

1D-Analysis1D-Analysis

3D-Analysis3D-Analysis

VI PowerVI Power

EnergyEnergy scalar quantity associated with a state scalar quantity associated with a state (or condition) of one or more objects (or condition) of one or more objects

II Kinetic energyKinetic energy

Energy associated with the state of motion of an objectEnergy associated with the state of motion of an object

2

2

1mvK

UnitsUnits 1 Joule = 1J = 1 kgm1 Joule = 1J = 1 kgm22ss22

II WorkII Work

Energy transferred ldquotordquo or ldquofromrdquo an Energy transferred ldquotordquo or ldquofromrdquo an object by means of a force acting on the object by means of a force acting on the object object

To To +W+WFrom From -W-W

- - Constant Constant forceforce

xx maF

dFWdFKKvvm

dvvmmaF

d

vvadavv

xxiff

fxxf

xxf

)(2

1

1)(

2

1

22

20

2

20

220

220

2

Work done by the force = Energy transfer due to the forceWork done by the force = Energy transfer due to the force

-To calculate the work done on an object by a force during To calculate the work done on an object by a force during a displacement we use only the force component alonga displacement we use only the force component along the objectrsquos displacement The force component the objectrsquos displacement The force component perpendicular to the displacement does zero workperpendicular to the displacement does zero work

dFFddFW x

cos

Assumptions Assumptions 1) F = constant force 1) F = constant force 2) Object is particle-like (rigid object all 2) Object is particle-like (rigid object all parts of the object must move together)parts of the object must move together)

090

90180

90

W

W

A force doesA force does +W+W when it has a vector component in the when it has a vector component in the same direction as the displacement andsame direction as the displacement and ndashWndashW when it has when it has a vector component in the opposite directiona vector component in the opposite direction W=0W=0 when it when it has no such vector componenthas no such vector component

Net work done by several forces = Sum of works done by Net work done by several forces = Sum of works done by individual forcesindividual forces CalculationCalculation 1) W1) Wnetnet= W= W11+W+W22+W+W33+hellip+hellip

2) F2) Fnet net W Wnetnet=F=Fnetnet d d

II Work-Kinetic Energy TheoremII Work-Kinetic Energy Theorem

WKKK if

Change in the kinetic energy of the Change in the kinetic energy of the particle particle == Net work done on the particle Net work done on the particle

III Work done by a constant forceIII Work done by a constant force

- - Gravitational forceGravitational force

cosmgdW

Rising objectRising object W= mgd cos180W= mgd cos180ordm ordm = -mgd= -mgd FFgg transfers transfers mgdmgd energyenergy from from the objectrsquos kinetic energy the objectrsquos kinetic energy

Falling down objectFalling down object W= mgd cos 0W= mgd cos 0ordm = +mgdordm = +mgd Fg transfers transfers mgd energyenergy to to the objectrsquos kinetic energy the objectrsquos kinetic energy

External applied force + Gravitational force External applied force + Gravitational force

gaif WWKKK

Object stationary before Object stationary before and after the liftand after the lift Wa+Wg=0

The applied force transfers the same The applied force transfers the same amount of energy to the object as the amount of energy to the object as the gravitational force transfers from the objectgravitational force transfers from the object

IV Work done by a variable forceIV Work done by a variable force

- - Spring forceSpring force

xkF

Hookersquos lawHookersquos law

k = spring constant = spring constant measures springrsquos stiffness measures springrsquos stiffness Units Nm

Hookersquos LawHookersquos Law

When When xx is positive is positive (spring is stretched)(spring is stretched) FF is negativeis negative

WhenWhen xx is is 00 (at the (at the equilibrium position)equilibrium position) F F isis 00

When When xx is negative is negative (spring is compressed)(spring is compressed) FF is positiveis positive

Hookersquos LawHookersquos Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the restoring forceis called the restoring force If the block is released it will If the block is released it will

oscillate back and forth between oscillate back and forth between ndashndashxx and and xx

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

Work ExampleWork Example

The normal forceThe normal force nn and and the gravitational forcethe gravitational force mgmg do no work on the objectdo no work on the object

cos cos θθ = cos 90deg= 0= cos 90deg= 0

The forceThe force FF does do work does do work on the objecton the object

More About WorkMore About Work The system and the environment must be The system and the environment must be

determined when dealing with workdetermined when dealing with work The environment does work on the systemThe environment does work on the system

(Work (Work byby the environment the environment onon the system) the system) The sign of the work depends on the direction ofThe sign of the work depends on the direction of

FF relative torelative to ΔΔrr Work is positive when projection ofWork is positive when projection of FF ontoonto ΔΔrr

is in the same direction as the displacementis in the same direction as the displacement Work is negative when the projection is in the Work is negative when the projection is in the

opposite directionopposite direction

Work Is An Energy TransferWork Is An Energy Transfer

bull This is important for a system approach This is important for a system approach to solving a problemto solving a problem

bull If the work is done on a system and it is If the work is done on a system and it is positive energy is transferred to the positive energy is transferred to the systemsystem

bull If the work done on the system is If the work done on the system is negative energy is transferred from the negative energy is transferred from the systemsystem

Work done by a constant forceWork done by a constant force

bull Work done on an object by a constant force Work done on an object by a constant force is defined to be the product of the magnitude of is defined to be the product of the magnitude of the the displacementdisplacement and the component of the and the component of the forceforce parallel to the displacement parallel to the displacement

bull WhereWhere FFIIII is the component of the forceis the component of the force FF parallel to the displacementparallel to the displacement dd

W = FW = FII II middotmiddot d d

WorkIn other words - In other words -

WhereWhere θθ is the angle betweenis the angle between FF and and dd

IfIf θθ is is gt 90gt 90oo work is negative A decelerating car has work is negative A decelerating car has negative work done on it by its enginenegative work done on it by its engine

The unit of work is called The unit of work is called Joule (J)Joule (J) 1J = 1 N1J = 1 Nmiddotmiddotmm

1J=kg∙m1J=kg∙m22ss22

d

F

W = F d cosW = F d cos

Scalar Product of Two VectorsScalar Product of Two Vectors

The scalar product The scalar product of two vectors is of two vectors is written aswritten as A A BB

It is also called It is also called the dot productthe dot product

A A bull bull B = A B cos B = A B cos θθ

θθ is the angle is the angle betweenbetween AA andand BB

Scalar ProductScalar Product

The scalar product is commutativeThe scalar product is commutative A A B = B B = B AA

The scalar product obeys the distributive law The scalar product obeys the distributive law of multiplicationof multiplication

A A (B + C) (B + C) == A A B + A B + A CC

Dot Products of Unit VectorsDot Products of Unit Vectors

bull Using component form withUsing component form with A A andand BB

0kjkiji

1kkjjii

zzyyxx

zyx

zyx

BABABABA

kBjBiBB

kAjAiAA

Work - on and byWork - on and by

A person pushes blockA person pushes block 30 m30 m along the along the ground by exerting force ofground by exerting force of 25 N25 N on the on the trolley How much work does the person do trolley How much work does the person do on the trolleyon the trolley

W = FW = Fd = 25N x 30m = 750 Nmd = 25N x 30m = 750 Nm

Trolley does -Trolley does -750 Nm750 Nm work on the personwork on the person

FFpptFFtrtr

d

A force acts on an object as A force acts on an object as the object moves in the the object moves in the xx direction from the origin direction from the origin to to xx = 500 m = 500 m Find the work Find the work W W done on the done on the object by the forceobject by the force

F 4x ˆ i 3yˆ j N

Mechanical EnergyMechanical Energy

Mechanical energy (energy associated Mechanical energy (energy associated

with with masses) can be thought of as having masses) can be thought of as having

two two components components kinetickinetic and and potentialpotential

ndashKinetic energy is energy of motionKinetic energy is energy of motion

ndashPotential energy is energy of positionPotential energy is energy of position

Kinetic EnergyKinetic Energy

bull In many situations energy can be In many situations energy can be considered asconsidered as ldquothe ability to do workrdquoldquothe ability to do workrdquo

bull Energy can exist in different formsEnergy can exist in different forms

bull Kinetic energy is the energy associated Kinetic energy is the energy associated with the with the motionmotion of an object of an object

bull A moving object can do work on A moving object can do work on another objectanother object

Eg hammer does work on nailEg hammer does work on nail

Kinetic EnergyKinetic Energy Consider an object with massConsider an object with mass mm moving in a straight line with moving in a straight line with initial velocityinitial velocity vvii To accelerate it uniformly to a speedTo accelerate it uniformly to a speed vvff a a constant net forceconstant net force FF is exerted on it parallel to motion over a is exerted on it parallel to motion over a distancedistance dd

Work done on objectWork done on object

W = F d = m a dW = F d = m a d (NII)(NII)SoSo

da

ada

if

iifif

2

2)(222

222

vv

vxxvv

dd

mW if

2

22 vv

Kinetic EnergyKinetic Energy

If we rearrange this we obtainIf we rearrange this we obtain

We define the quantityWe define the quantity frac12mvfrac12mv22 to be the to be the translational translational kinetic energykinetic energy (KE) of the object (KE) of the object

This is the This is the lsquolsquoWork-Energy TheoremWork-Energy Theoremrsquorsquo

ldquoldquoThe net The net workwork done on an object is equal to done on an object is equal to its its change in kinetic energychange in kinetic energyrdquordquo

22

2

1

2

1if mmW vv

W = KE

The Work-Energy TheoremThe Work-Energy Theorem

bull NoteNotendash The net work done on an object is the work The net work done on an object is the work

done by the net forcedone by the net forcendash Units of energy and work must be the Units of energy and work must be the

same (same (JJ))

W = W = KEKE

II Kinetic energyKinetic energy

IIII WorkWork

IIIIII Work - Kinetic energy theoremWork - Kinetic energy theorem

IVIV Work done by a constant forceWork done by a constant force

- Gravitational force- Gravitational force

V Work done by a variable forceV Work done by a variable force

- Spring force- Spring force

1D-Analysis1D-Analysis

3D-Analysis3D-Analysis

VI PowerVI Power

EnergyEnergy scalar quantity associated with a state scalar quantity associated with a state (or condition) of one or more objects (or condition) of one or more objects

II Kinetic energyKinetic energy

Energy associated with the state of motion of an objectEnergy associated with the state of motion of an object

2

2

1mvK

UnitsUnits 1 Joule = 1J = 1 kgm1 Joule = 1J = 1 kgm22ss22

II WorkII Work

Energy transferred ldquotordquo or ldquofromrdquo an Energy transferred ldquotordquo or ldquofromrdquo an object by means of a force acting on the object by means of a force acting on the object object

To To +W+WFrom From -W-W

- - Constant Constant forceforce

xx maF

dFWdFKKvvm

dvvmmaF

d

vvadavv

xxiff

fxxf

xxf

)(2

1

1)(

2

1

22

20

2

20

220

220

2

Work done by the force = Energy transfer due to the forceWork done by the force = Energy transfer due to the force

-To calculate the work done on an object by a force during To calculate the work done on an object by a force during a displacement we use only the force component alonga displacement we use only the force component along the objectrsquos displacement The force component the objectrsquos displacement The force component perpendicular to the displacement does zero workperpendicular to the displacement does zero work

dFFddFW x

cos

Assumptions Assumptions 1) F = constant force 1) F = constant force 2) Object is particle-like (rigid object all 2) Object is particle-like (rigid object all parts of the object must move together)parts of the object must move together)

090

90180

90

W

W

A force doesA force does +W+W when it has a vector component in the when it has a vector component in the same direction as the displacement andsame direction as the displacement and ndashWndashW when it has when it has a vector component in the opposite directiona vector component in the opposite direction W=0W=0 when it when it has no such vector componenthas no such vector component

Net work done by several forces = Sum of works done by Net work done by several forces = Sum of works done by individual forcesindividual forces CalculationCalculation 1) W1) Wnetnet= W= W11+W+W22+W+W33+hellip+hellip

2) F2) Fnet net W Wnetnet=F=Fnetnet d d

II Work-Kinetic Energy TheoremII Work-Kinetic Energy Theorem

WKKK if

Change in the kinetic energy of the Change in the kinetic energy of the particle particle == Net work done on the particle Net work done on the particle

III Work done by a constant forceIII Work done by a constant force

- - Gravitational forceGravitational force

cosmgdW

Rising objectRising object W= mgd cos180W= mgd cos180ordm ordm = -mgd= -mgd FFgg transfers transfers mgdmgd energyenergy from from the objectrsquos kinetic energy the objectrsquos kinetic energy

Falling down objectFalling down object W= mgd cos 0W= mgd cos 0ordm = +mgdordm = +mgd Fg transfers transfers mgd energyenergy to to the objectrsquos kinetic energy the objectrsquos kinetic energy

External applied force + Gravitational force External applied force + Gravitational force

gaif WWKKK

Object stationary before Object stationary before and after the liftand after the lift Wa+Wg=0

The applied force transfers the same The applied force transfers the same amount of energy to the object as the amount of energy to the object as the gravitational force transfers from the objectgravitational force transfers from the object

IV Work done by a variable forceIV Work done by a variable force

- - Spring forceSpring force

xkF

Hookersquos lawHookersquos law

k = spring constant = spring constant measures springrsquos stiffness measures springrsquos stiffness Units Nm

Hookersquos LawHookersquos Law

When When xx is positive is positive (spring is stretched)(spring is stretched) FF is negativeis negative

WhenWhen xx is is 00 (at the (at the equilibrium position)equilibrium position) F F isis 00

When When xx is negative is negative (spring is compressed)(spring is compressed) FF is positiveis positive

Hookersquos LawHookersquos Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the restoring forceis called the restoring force If the block is released it will If the block is released it will

oscillate back and forth between oscillate back and forth between ndashndashxx and and xx

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

More About WorkMore About Work The system and the environment must be The system and the environment must be

determined when dealing with workdetermined when dealing with work The environment does work on the systemThe environment does work on the system

(Work (Work byby the environment the environment onon the system) the system) The sign of the work depends on the direction ofThe sign of the work depends on the direction of

FF relative torelative to ΔΔrr Work is positive when projection ofWork is positive when projection of FF ontoonto ΔΔrr

is in the same direction as the displacementis in the same direction as the displacement Work is negative when the projection is in the Work is negative when the projection is in the

opposite directionopposite direction

Work Is An Energy TransferWork Is An Energy Transfer

bull This is important for a system approach This is important for a system approach to solving a problemto solving a problem

bull If the work is done on a system and it is If the work is done on a system and it is positive energy is transferred to the positive energy is transferred to the systemsystem

bull If the work done on the system is If the work done on the system is negative energy is transferred from the negative energy is transferred from the systemsystem

Work done by a constant forceWork done by a constant force

bull Work done on an object by a constant force Work done on an object by a constant force is defined to be the product of the magnitude of is defined to be the product of the magnitude of the the displacementdisplacement and the component of the and the component of the forceforce parallel to the displacement parallel to the displacement

bull WhereWhere FFIIII is the component of the forceis the component of the force FF parallel to the displacementparallel to the displacement dd

W = FW = FII II middotmiddot d d

WorkIn other words - In other words -

WhereWhere θθ is the angle betweenis the angle between FF and and dd

IfIf θθ is is gt 90gt 90oo work is negative A decelerating car has work is negative A decelerating car has negative work done on it by its enginenegative work done on it by its engine

The unit of work is called The unit of work is called Joule (J)Joule (J) 1J = 1 N1J = 1 Nmiddotmiddotmm

1J=kg∙m1J=kg∙m22ss22

d

F

W = F d cosW = F d cos

Scalar Product of Two VectorsScalar Product of Two Vectors

The scalar product The scalar product of two vectors is of two vectors is written aswritten as A A BB

It is also called It is also called the dot productthe dot product

A A bull bull B = A B cos B = A B cos θθ

θθ is the angle is the angle betweenbetween AA andand BB

Scalar ProductScalar Product

The scalar product is commutativeThe scalar product is commutative A A B = B B = B AA

The scalar product obeys the distributive law The scalar product obeys the distributive law of multiplicationof multiplication

A A (B + C) (B + C) == A A B + A B + A CC

Dot Products of Unit VectorsDot Products of Unit Vectors

bull Using component form withUsing component form with A A andand BB

0kjkiji

1kkjjii

zzyyxx

zyx

zyx

BABABABA

kBjBiBB

kAjAiAA

Work - on and byWork - on and by

A person pushes blockA person pushes block 30 m30 m along the along the ground by exerting force ofground by exerting force of 25 N25 N on the on the trolley How much work does the person do trolley How much work does the person do on the trolleyon the trolley

W = FW = Fd = 25N x 30m = 750 Nmd = 25N x 30m = 750 Nm

Trolley does -Trolley does -750 Nm750 Nm work on the personwork on the person

FFpptFFtrtr

d

A force acts on an object as A force acts on an object as the object moves in the the object moves in the xx direction from the origin direction from the origin to to xx = 500 m = 500 m Find the work Find the work W W done on the done on the object by the forceobject by the force

F 4x ˆ i 3yˆ j N

Mechanical EnergyMechanical Energy

Mechanical energy (energy associated Mechanical energy (energy associated

with with masses) can be thought of as having masses) can be thought of as having

two two components components kinetickinetic and and potentialpotential

ndashKinetic energy is energy of motionKinetic energy is energy of motion

ndashPotential energy is energy of positionPotential energy is energy of position

Kinetic EnergyKinetic Energy

bull In many situations energy can be In many situations energy can be considered asconsidered as ldquothe ability to do workrdquoldquothe ability to do workrdquo

bull Energy can exist in different formsEnergy can exist in different forms

bull Kinetic energy is the energy associated Kinetic energy is the energy associated with the with the motionmotion of an object of an object

bull A moving object can do work on A moving object can do work on another objectanother object

Eg hammer does work on nailEg hammer does work on nail

Kinetic EnergyKinetic Energy Consider an object with massConsider an object with mass mm moving in a straight line with moving in a straight line with initial velocityinitial velocity vvii To accelerate it uniformly to a speedTo accelerate it uniformly to a speed vvff a a constant net forceconstant net force FF is exerted on it parallel to motion over a is exerted on it parallel to motion over a distancedistance dd

Work done on objectWork done on object

W = F d = m a dW = F d = m a d (NII)(NII)SoSo

da

ada

if

iifif

2

2)(222

222

vv

vxxvv

dd

mW if

2

22 vv

Kinetic EnergyKinetic Energy

If we rearrange this we obtainIf we rearrange this we obtain

We define the quantityWe define the quantity frac12mvfrac12mv22 to be the to be the translational translational kinetic energykinetic energy (KE) of the object (KE) of the object

This is the This is the lsquolsquoWork-Energy TheoremWork-Energy Theoremrsquorsquo

ldquoldquoThe net The net workwork done on an object is equal to done on an object is equal to its its change in kinetic energychange in kinetic energyrdquordquo

22

2

1

2

1if mmW vv

W = KE

The Work-Energy TheoremThe Work-Energy Theorem

bull NoteNotendash The net work done on an object is the work The net work done on an object is the work

done by the net forcedone by the net forcendash Units of energy and work must be the Units of energy and work must be the

same (same (JJ))

W = W = KEKE

II Kinetic energyKinetic energy

IIII WorkWork

IIIIII Work - Kinetic energy theoremWork - Kinetic energy theorem

IVIV Work done by a constant forceWork done by a constant force

- Gravitational force- Gravitational force

V Work done by a variable forceV Work done by a variable force

- Spring force- Spring force

1D-Analysis1D-Analysis

3D-Analysis3D-Analysis

VI PowerVI Power

EnergyEnergy scalar quantity associated with a state scalar quantity associated with a state (or condition) of one or more objects (or condition) of one or more objects

II Kinetic energyKinetic energy

Energy associated with the state of motion of an objectEnergy associated with the state of motion of an object

2

2

1mvK

UnitsUnits 1 Joule = 1J = 1 kgm1 Joule = 1J = 1 kgm22ss22

II WorkII Work

Energy transferred ldquotordquo or ldquofromrdquo an Energy transferred ldquotordquo or ldquofromrdquo an object by means of a force acting on the object by means of a force acting on the object object

To To +W+WFrom From -W-W

- - Constant Constant forceforce

xx maF

dFWdFKKvvm

dvvmmaF

d

vvadavv

xxiff

fxxf

xxf

)(2

1

1)(

2

1

22

20

2

20

220

220

2

Work done by the force = Energy transfer due to the forceWork done by the force = Energy transfer due to the force

-To calculate the work done on an object by a force during To calculate the work done on an object by a force during a displacement we use only the force component alonga displacement we use only the force component along the objectrsquos displacement The force component the objectrsquos displacement The force component perpendicular to the displacement does zero workperpendicular to the displacement does zero work

dFFddFW x

cos

Assumptions Assumptions 1) F = constant force 1) F = constant force 2) Object is particle-like (rigid object all 2) Object is particle-like (rigid object all parts of the object must move together)parts of the object must move together)

090

90180

90

W

W

A force doesA force does +W+W when it has a vector component in the when it has a vector component in the same direction as the displacement andsame direction as the displacement and ndashWndashW when it has when it has a vector component in the opposite directiona vector component in the opposite direction W=0W=0 when it when it has no such vector componenthas no such vector component

Net work done by several forces = Sum of works done by Net work done by several forces = Sum of works done by individual forcesindividual forces CalculationCalculation 1) W1) Wnetnet= W= W11+W+W22+W+W33+hellip+hellip

2) F2) Fnet net W Wnetnet=F=Fnetnet d d

II Work-Kinetic Energy TheoremII Work-Kinetic Energy Theorem

WKKK if

Change in the kinetic energy of the Change in the kinetic energy of the particle particle == Net work done on the particle Net work done on the particle

III Work done by a constant forceIII Work done by a constant force

- - Gravitational forceGravitational force

cosmgdW

Rising objectRising object W= mgd cos180W= mgd cos180ordm ordm = -mgd= -mgd FFgg transfers transfers mgdmgd energyenergy from from the objectrsquos kinetic energy the objectrsquos kinetic energy

Falling down objectFalling down object W= mgd cos 0W= mgd cos 0ordm = +mgdordm = +mgd Fg transfers transfers mgd energyenergy to to the objectrsquos kinetic energy the objectrsquos kinetic energy

External applied force + Gravitational force External applied force + Gravitational force

gaif WWKKK

Object stationary before Object stationary before and after the liftand after the lift Wa+Wg=0

The applied force transfers the same The applied force transfers the same amount of energy to the object as the amount of energy to the object as the gravitational force transfers from the objectgravitational force transfers from the object

IV Work done by a variable forceIV Work done by a variable force

- - Spring forceSpring force

xkF

Hookersquos lawHookersquos law

k = spring constant = spring constant measures springrsquos stiffness measures springrsquos stiffness Units Nm

Hookersquos LawHookersquos Law

When When xx is positive is positive (spring is stretched)(spring is stretched) FF is negativeis negative

WhenWhen xx is is 00 (at the (at the equilibrium position)equilibrium position) F F isis 00

When When xx is negative is negative (spring is compressed)(spring is compressed) FF is positiveis positive

Hookersquos LawHookersquos Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the restoring forceis called the restoring force If the block is released it will If the block is released it will

oscillate back and forth between oscillate back and forth between ndashndashxx and and xx

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

Work Is An Energy TransferWork Is An Energy Transfer

bull This is important for a system approach This is important for a system approach to solving a problemto solving a problem

bull If the work is done on a system and it is If the work is done on a system and it is positive energy is transferred to the positive energy is transferred to the systemsystem

bull If the work done on the system is If the work done on the system is negative energy is transferred from the negative energy is transferred from the systemsystem

Work done by a constant forceWork done by a constant force

bull Work done on an object by a constant force Work done on an object by a constant force is defined to be the product of the magnitude of is defined to be the product of the magnitude of the the displacementdisplacement and the component of the and the component of the forceforce parallel to the displacement parallel to the displacement

bull WhereWhere FFIIII is the component of the forceis the component of the force FF parallel to the displacementparallel to the displacement dd

W = FW = FII II middotmiddot d d

WorkIn other words - In other words -

WhereWhere θθ is the angle betweenis the angle between FF and and dd

IfIf θθ is is gt 90gt 90oo work is negative A decelerating car has work is negative A decelerating car has negative work done on it by its enginenegative work done on it by its engine

The unit of work is called The unit of work is called Joule (J)Joule (J) 1J = 1 N1J = 1 Nmiddotmiddotmm

1J=kg∙m1J=kg∙m22ss22

d

F

W = F d cosW = F d cos

Scalar Product of Two VectorsScalar Product of Two Vectors

The scalar product The scalar product of two vectors is of two vectors is written aswritten as A A BB

It is also called It is also called the dot productthe dot product

A A bull bull B = A B cos B = A B cos θθ

θθ is the angle is the angle betweenbetween AA andand BB

Scalar ProductScalar Product

The scalar product is commutativeThe scalar product is commutative A A B = B B = B AA

The scalar product obeys the distributive law The scalar product obeys the distributive law of multiplicationof multiplication

A A (B + C) (B + C) == A A B + A B + A CC

Dot Products of Unit VectorsDot Products of Unit Vectors

bull Using component form withUsing component form with A A andand BB

0kjkiji

1kkjjii

zzyyxx

zyx

zyx

BABABABA

kBjBiBB

kAjAiAA

Work - on and byWork - on and by

A person pushes blockA person pushes block 30 m30 m along the along the ground by exerting force ofground by exerting force of 25 N25 N on the on the trolley How much work does the person do trolley How much work does the person do on the trolleyon the trolley

W = FW = Fd = 25N x 30m = 750 Nmd = 25N x 30m = 750 Nm

Trolley does -Trolley does -750 Nm750 Nm work on the personwork on the person

FFpptFFtrtr

d

A force acts on an object as A force acts on an object as the object moves in the the object moves in the xx direction from the origin direction from the origin to to xx = 500 m = 500 m Find the work Find the work W W done on the done on the object by the forceobject by the force

F 4x ˆ i 3yˆ j N

Mechanical EnergyMechanical Energy

Mechanical energy (energy associated Mechanical energy (energy associated

with with masses) can be thought of as having masses) can be thought of as having

two two components components kinetickinetic and and potentialpotential

ndashKinetic energy is energy of motionKinetic energy is energy of motion

ndashPotential energy is energy of positionPotential energy is energy of position

Kinetic EnergyKinetic Energy

bull In many situations energy can be In many situations energy can be considered asconsidered as ldquothe ability to do workrdquoldquothe ability to do workrdquo

bull Energy can exist in different formsEnergy can exist in different forms

bull Kinetic energy is the energy associated Kinetic energy is the energy associated with the with the motionmotion of an object of an object

bull A moving object can do work on A moving object can do work on another objectanother object

Eg hammer does work on nailEg hammer does work on nail

Kinetic EnergyKinetic Energy Consider an object with massConsider an object with mass mm moving in a straight line with moving in a straight line with initial velocityinitial velocity vvii To accelerate it uniformly to a speedTo accelerate it uniformly to a speed vvff a a constant net forceconstant net force FF is exerted on it parallel to motion over a is exerted on it parallel to motion over a distancedistance dd

Work done on objectWork done on object

W = F d = m a dW = F d = m a d (NII)(NII)SoSo

da

ada

if

iifif

2

2)(222

222

vv

vxxvv

dd

mW if

2

22 vv

Kinetic EnergyKinetic Energy

If we rearrange this we obtainIf we rearrange this we obtain

We define the quantityWe define the quantity frac12mvfrac12mv22 to be the to be the translational translational kinetic energykinetic energy (KE) of the object (KE) of the object

This is the This is the lsquolsquoWork-Energy TheoremWork-Energy Theoremrsquorsquo

ldquoldquoThe net The net workwork done on an object is equal to done on an object is equal to its its change in kinetic energychange in kinetic energyrdquordquo

22

2

1

2

1if mmW vv

W = KE

The Work-Energy TheoremThe Work-Energy Theorem

bull NoteNotendash The net work done on an object is the work The net work done on an object is the work

done by the net forcedone by the net forcendash Units of energy and work must be the Units of energy and work must be the

same (same (JJ))

W = W = KEKE

II Kinetic energyKinetic energy

IIII WorkWork

IIIIII Work - Kinetic energy theoremWork - Kinetic energy theorem

IVIV Work done by a constant forceWork done by a constant force

- Gravitational force- Gravitational force

V Work done by a variable forceV Work done by a variable force

- Spring force- Spring force

1D-Analysis1D-Analysis

3D-Analysis3D-Analysis

VI PowerVI Power

EnergyEnergy scalar quantity associated with a state scalar quantity associated with a state (or condition) of one or more objects (or condition) of one or more objects

II Kinetic energyKinetic energy

Energy associated with the state of motion of an objectEnergy associated with the state of motion of an object

2

2

1mvK

UnitsUnits 1 Joule = 1J = 1 kgm1 Joule = 1J = 1 kgm22ss22

II WorkII Work

Energy transferred ldquotordquo or ldquofromrdquo an Energy transferred ldquotordquo or ldquofromrdquo an object by means of a force acting on the object by means of a force acting on the object object

To To +W+WFrom From -W-W

- - Constant Constant forceforce

xx maF

dFWdFKKvvm

dvvmmaF

d

vvadavv

xxiff

fxxf

xxf

)(2

1

1)(

2

1

22

20

2

20

220

220

2

Work done by the force = Energy transfer due to the forceWork done by the force = Energy transfer due to the force

-To calculate the work done on an object by a force during To calculate the work done on an object by a force during a displacement we use only the force component alonga displacement we use only the force component along the objectrsquos displacement The force component the objectrsquos displacement The force component perpendicular to the displacement does zero workperpendicular to the displacement does zero work

dFFddFW x

cos

Assumptions Assumptions 1) F = constant force 1) F = constant force 2) Object is particle-like (rigid object all 2) Object is particle-like (rigid object all parts of the object must move together)parts of the object must move together)

090

90180

90

W

W

A force doesA force does +W+W when it has a vector component in the when it has a vector component in the same direction as the displacement andsame direction as the displacement and ndashWndashW when it has when it has a vector component in the opposite directiona vector component in the opposite direction W=0W=0 when it when it has no such vector componenthas no such vector component

Net work done by several forces = Sum of works done by Net work done by several forces = Sum of works done by individual forcesindividual forces CalculationCalculation 1) W1) Wnetnet= W= W11+W+W22+W+W33+hellip+hellip

2) F2) Fnet net W Wnetnet=F=Fnetnet d d

II Work-Kinetic Energy TheoremII Work-Kinetic Energy Theorem

WKKK if

Change in the kinetic energy of the Change in the kinetic energy of the particle particle == Net work done on the particle Net work done on the particle

III Work done by a constant forceIII Work done by a constant force

- - Gravitational forceGravitational force

cosmgdW

Rising objectRising object W= mgd cos180W= mgd cos180ordm ordm = -mgd= -mgd FFgg transfers transfers mgdmgd energyenergy from from the objectrsquos kinetic energy the objectrsquos kinetic energy

Falling down objectFalling down object W= mgd cos 0W= mgd cos 0ordm = +mgdordm = +mgd Fg transfers transfers mgd energyenergy to to the objectrsquos kinetic energy the objectrsquos kinetic energy

External applied force + Gravitational force External applied force + Gravitational force

gaif WWKKK

Object stationary before Object stationary before and after the liftand after the lift Wa+Wg=0

The applied force transfers the same The applied force transfers the same amount of energy to the object as the amount of energy to the object as the gravitational force transfers from the objectgravitational force transfers from the object

IV Work done by a variable forceIV Work done by a variable force

- - Spring forceSpring force

xkF

Hookersquos lawHookersquos law

k = spring constant = spring constant measures springrsquos stiffness measures springrsquos stiffness Units Nm

Hookersquos LawHookersquos Law

When When xx is positive is positive (spring is stretched)(spring is stretched) FF is negativeis negative

WhenWhen xx is is 00 (at the (at the equilibrium position)equilibrium position) F F isis 00

When When xx is negative is negative (spring is compressed)(spring is compressed) FF is positiveis positive

Hookersquos LawHookersquos Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the restoring forceis called the restoring force If the block is released it will If the block is released it will

oscillate back and forth between oscillate back and forth between ndashndashxx and and xx

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

Work done by a constant forceWork done by a constant force

bull Work done on an object by a constant force Work done on an object by a constant force is defined to be the product of the magnitude of is defined to be the product of the magnitude of the the displacementdisplacement and the component of the and the component of the forceforce parallel to the displacement parallel to the displacement

bull WhereWhere FFIIII is the component of the forceis the component of the force FF parallel to the displacementparallel to the displacement dd

W = FW = FII II middotmiddot d d

WorkIn other words - In other words -

WhereWhere θθ is the angle betweenis the angle between FF and and dd

IfIf θθ is is gt 90gt 90oo work is negative A decelerating car has work is negative A decelerating car has negative work done on it by its enginenegative work done on it by its engine

The unit of work is called The unit of work is called Joule (J)Joule (J) 1J = 1 N1J = 1 Nmiddotmiddotmm

1J=kg∙m1J=kg∙m22ss22

d

F

W = F d cosW = F d cos

Scalar Product of Two VectorsScalar Product of Two Vectors

The scalar product The scalar product of two vectors is of two vectors is written aswritten as A A BB

It is also called It is also called the dot productthe dot product

A A bull bull B = A B cos B = A B cos θθ

θθ is the angle is the angle betweenbetween AA andand BB

Scalar ProductScalar Product

The scalar product is commutativeThe scalar product is commutative A A B = B B = B AA

The scalar product obeys the distributive law The scalar product obeys the distributive law of multiplicationof multiplication

A A (B + C) (B + C) == A A B + A B + A CC

Dot Products of Unit VectorsDot Products of Unit Vectors

bull Using component form withUsing component form with A A andand BB

0kjkiji

1kkjjii

zzyyxx

zyx

zyx

BABABABA

kBjBiBB

kAjAiAA

Work - on and byWork - on and by

A person pushes blockA person pushes block 30 m30 m along the along the ground by exerting force ofground by exerting force of 25 N25 N on the on the trolley How much work does the person do trolley How much work does the person do on the trolleyon the trolley

W = FW = Fd = 25N x 30m = 750 Nmd = 25N x 30m = 750 Nm

Trolley does -Trolley does -750 Nm750 Nm work on the personwork on the person

FFpptFFtrtr

d

A force acts on an object as A force acts on an object as the object moves in the the object moves in the xx direction from the origin direction from the origin to to xx = 500 m = 500 m Find the work Find the work W W done on the done on the object by the forceobject by the force

F 4x ˆ i 3yˆ j N

Mechanical EnergyMechanical Energy

Mechanical energy (energy associated Mechanical energy (energy associated

with with masses) can be thought of as having masses) can be thought of as having

two two components components kinetickinetic and and potentialpotential

ndashKinetic energy is energy of motionKinetic energy is energy of motion

ndashPotential energy is energy of positionPotential energy is energy of position

Kinetic EnergyKinetic Energy

bull In many situations energy can be In many situations energy can be considered asconsidered as ldquothe ability to do workrdquoldquothe ability to do workrdquo

bull Energy can exist in different formsEnergy can exist in different forms

bull Kinetic energy is the energy associated Kinetic energy is the energy associated with the with the motionmotion of an object of an object

bull A moving object can do work on A moving object can do work on another objectanother object

Eg hammer does work on nailEg hammer does work on nail

Kinetic EnergyKinetic Energy Consider an object with massConsider an object with mass mm moving in a straight line with moving in a straight line with initial velocityinitial velocity vvii To accelerate it uniformly to a speedTo accelerate it uniformly to a speed vvff a a constant net forceconstant net force FF is exerted on it parallel to motion over a is exerted on it parallel to motion over a distancedistance dd

Work done on objectWork done on object

W = F d = m a dW = F d = m a d (NII)(NII)SoSo

da

ada

if

iifif

2

2)(222

222

vv

vxxvv

dd

mW if

2

22 vv

Kinetic EnergyKinetic Energy

If we rearrange this we obtainIf we rearrange this we obtain

We define the quantityWe define the quantity frac12mvfrac12mv22 to be the to be the translational translational kinetic energykinetic energy (KE) of the object (KE) of the object

This is the This is the lsquolsquoWork-Energy TheoremWork-Energy Theoremrsquorsquo

ldquoldquoThe net The net workwork done on an object is equal to done on an object is equal to its its change in kinetic energychange in kinetic energyrdquordquo

22

2

1

2

1if mmW vv

W = KE

The Work-Energy TheoremThe Work-Energy Theorem

bull NoteNotendash The net work done on an object is the work The net work done on an object is the work

done by the net forcedone by the net forcendash Units of energy and work must be the Units of energy and work must be the

same (same (JJ))

W = W = KEKE

II Kinetic energyKinetic energy

IIII WorkWork

IIIIII Work - Kinetic energy theoremWork - Kinetic energy theorem

IVIV Work done by a constant forceWork done by a constant force

- Gravitational force- Gravitational force

V Work done by a variable forceV Work done by a variable force

- Spring force- Spring force

1D-Analysis1D-Analysis

3D-Analysis3D-Analysis

VI PowerVI Power

EnergyEnergy scalar quantity associated with a state scalar quantity associated with a state (or condition) of one or more objects (or condition) of one or more objects

II Kinetic energyKinetic energy

Energy associated with the state of motion of an objectEnergy associated with the state of motion of an object

2

2

1mvK

UnitsUnits 1 Joule = 1J = 1 kgm1 Joule = 1J = 1 kgm22ss22

II WorkII Work

Energy transferred ldquotordquo or ldquofromrdquo an Energy transferred ldquotordquo or ldquofromrdquo an object by means of a force acting on the object by means of a force acting on the object object

To To +W+WFrom From -W-W

- - Constant Constant forceforce

xx maF

dFWdFKKvvm

dvvmmaF

d

vvadavv

xxiff

fxxf

xxf

)(2

1

1)(

2

1

22

20

2

20

220

220

2

Work done by the force = Energy transfer due to the forceWork done by the force = Energy transfer due to the force

-To calculate the work done on an object by a force during To calculate the work done on an object by a force during a displacement we use only the force component alonga displacement we use only the force component along the objectrsquos displacement The force component the objectrsquos displacement The force component perpendicular to the displacement does zero workperpendicular to the displacement does zero work

dFFddFW x

cos

Assumptions Assumptions 1) F = constant force 1) F = constant force 2) Object is particle-like (rigid object all 2) Object is particle-like (rigid object all parts of the object must move together)parts of the object must move together)

090

90180

90

W

W

A force doesA force does +W+W when it has a vector component in the when it has a vector component in the same direction as the displacement andsame direction as the displacement and ndashWndashW when it has when it has a vector component in the opposite directiona vector component in the opposite direction W=0W=0 when it when it has no such vector componenthas no such vector component

Net work done by several forces = Sum of works done by Net work done by several forces = Sum of works done by individual forcesindividual forces CalculationCalculation 1) W1) Wnetnet= W= W11+W+W22+W+W33+hellip+hellip

2) F2) Fnet net W Wnetnet=F=Fnetnet d d

II Work-Kinetic Energy TheoremII Work-Kinetic Energy Theorem

WKKK if

Change in the kinetic energy of the Change in the kinetic energy of the particle particle == Net work done on the particle Net work done on the particle

III Work done by a constant forceIII Work done by a constant force

- - Gravitational forceGravitational force

cosmgdW

Rising objectRising object W= mgd cos180W= mgd cos180ordm ordm = -mgd= -mgd FFgg transfers transfers mgdmgd energyenergy from from the objectrsquos kinetic energy the objectrsquos kinetic energy

Falling down objectFalling down object W= mgd cos 0W= mgd cos 0ordm = +mgdordm = +mgd Fg transfers transfers mgd energyenergy to to the objectrsquos kinetic energy the objectrsquos kinetic energy

External applied force + Gravitational force External applied force + Gravitational force

gaif WWKKK

Object stationary before Object stationary before and after the liftand after the lift Wa+Wg=0

The applied force transfers the same The applied force transfers the same amount of energy to the object as the amount of energy to the object as the gravitational force transfers from the objectgravitational force transfers from the object

IV Work done by a variable forceIV Work done by a variable force

- - Spring forceSpring force

xkF

Hookersquos lawHookersquos law

k = spring constant = spring constant measures springrsquos stiffness measures springrsquos stiffness Units Nm

Hookersquos LawHookersquos Law

When When xx is positive is positive (spring is stretched)(spring is stretched) FF is negativeis negative

WhenWhen xx is is 00 (at the (at the equilibrium position)equilibrium position) F F isis 00

When When xx is negative is negative (spring is compressed)(spring is compressed) FF is positiveis positive

Hookersquos LawHookersquos Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the restoring forceis called the restoring force If the block is released it will If the block is released it will

oscillate back and forth between oscillate back and forth between ndashndashxx and and xx

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

WorkIn other words - In other words -

WhereWhere θθ is the angle betweenis the angle between FF and and dd

IfIf θθ is is gt 90gt 90oo work is negative A decelerating car has work is negative A decelerating car has negative work done on it by its enginenegative work done on it by its engine

The unit of work is called The unit of work is called Joule (J)Joule (J) 1J = 1 N1J = 1 Nmiddotmiddotmm

1J=kg∙m1J=kg∙m22ss22

d

F

W = F d cosW = F d cos

Scalar Product of Two VectorsScalar Product of Two Vectors

The scalar product The scalar product of two vectors is of two vectors is written aswritten as A A BB

It is also called It is also called the dot productthe dot product

A A bull bull B = A B cos B = A B cos θθ

θθ is the angle is the angle betweenbetween AA andand BB

Scalar ProductScalar Product

The scalar product is commutativeThe scalar product is commutative A A B = B B = B AA

The scalar product obeys the distributive law The scalar product obeys the distributive law of multiplicationof multiplication

A A (B + C) (B + C) == A A B + A B + A CC

Dot Products of Unit VectorsDot Products of Unit Vectors

bull Using component form withUsing component form with A A andand BB

0kjkiji

1kkjjii

zzyyxx

zyx

zyx

BABABABA

kBjBiBB

kAjAiAA

Work - on and byWork - on and by

A person pushes blockA person pushes block 30 m30 m along the along the ground by exerting force ofground by exerting force of 25 N25 N on the on the trolley How much work does the person do trolley How much work does the person do on the trolleyon the trolley

W = FW = Fd = 25N x 30m = 750 Nmd = 25N x 30m = 750 Nm

Trolley does -Trolley does -750 Nm750 Nm work on the personwork on the person

FFpptFFtrtr

d

A force acts on an object as A force acts on an object as the object moves in the the object moves in the xx direction from the origin direction from the origin to to xx = 500 m = 500 m Find the work Find the work W W done on the done on the object by the forceobject by the force

F 4x ˆ i 3yˆ j N

Mechanical EnergyMechanical Energy

Mechanical energy (energy associated Mechanical energy (energy associated

with with masses) can be thought of as having masses) can be thought of as having

two two components components kinetickinetic and and potentialpotential

ndashKinetic energy is energy of motionKinetic energy is energy of motion

ndashPotential energy is energy of positionPotential energy is energy of position

Kinetic EnergyKinetic Energy

bull In many situations energy can be In many situations energy can be considered asconsidered as ldquothe ability to do workrdquoldquothe ability to do workrdquo

bull Energy can exist in different formsEnergy can exist in different forms

bull Kinetic energy is the energy associated Kinetic energy is the energy associated with the with the motionmotion of an object of an object

bull A moving object can do work on A moving object can do work on another objectanother object

Eg hammer does work on nailEg hammer does work on nail

Kinetic EnergyKinetic Energy Consider an object with massConsider an object with mass mm moving in a straight line with moving in a straight line with initial velocityinitial velocity vvii To accelerate it uniformly to a speedTo accelerate it uniformly to a speed vvff a a constant net forceconstant net force FF is exerted on it parallel to motion over a is exerted on it parallel to motion over a distancedistance dd

Work done on objectWork done on object

W = F d = m a dW = F d = m a d (NII)(NII)SoSo

da

ada

if

iifif

2

2)(222

222

vv

vxxvv

dd

mW if

2

22 vv

Kinetic EnergyKinetic Energy

If we rearrange this we obtainIf we rearrange this we obtain

We define the quantityWe define the quantity frac12mvfrac12mv22 to be the to be the translational translational kinetic energykinetic energy (KE) of the object (KE) of the object

This is the This is the lsquolsquoWork-Energy TheoremWork-Energy Theoremrsquorsquo

ldquoldquoThe net The net workwork done on an object is equal to done on an object is equal to its its change in kinetic energychange in kinetic energyrdquordquo

22

2

1

2

1if mmW vv

W = KE

The Work-Energy TheoremThe Work-Energy Theorem

bull NoteNotendash The net work done on an object is the work The net work done on an object is the work

done by the net forcedone by the net forcendash Units of energy and work must be the Units of energy and work must be the

same (same (JJ))

W = W = KEKE

II Kinetic energyKinetic energy

IIII WorkWork

IIIIII Work - Kinetic energy theoremWork - Kinetic energy theorem

IVIV Work done by a constant forceWork done by a constant force

- Gravitational force- Gravitational force

V Work done by a variable forceV Work done by a variable force

- Spring force- Spring force

1D-Analysis1D-Analysis

3D-Analysis3D-Analysis

VI PowerVI Power

EnergyEnergy scalar quantity associated with a state scalar quantity associated with a state (or condition) of one or more objects (or condition) of one or more objects

II Kinetic energyKinetic energy

Energy associated with the state of motion of an objectEnergy associated with the state of motion of an object

2

2

1mvK

UnitsUnits 1 Joule = 1J = 1 kgm1 Joule = 1J = 1 kgm22ss22

II WorkII Work

Energy transferred ldquotordquo or ldquofromrdquo an Energy transferred ldquotordquo or ldquofromrdquo an object by means of a force acting on the object by means of a force acting on the object object

To To +W+WFrom From -W-W

- - Constant Constant forceforce

xx maF

dFWdFKKvvm

dvvmmaF

d

vvadavv

xxiff

fxxf

xxf

)(2

1

1)(

2

1

22

20

2

20

220

220

2

Work done by the force = Energy transfer due to the forceWork done by the force = Energy transfer due to the force

-To calculate the work done on an object by a force during To calculate the work done on an object by a force during a displacement we use only the force component alonga displacement we use only the force component along the objectrsquos displacement The force component the objectrsquos displacement The force component perpendicular to the displacement does zero workperpendicular to the displacement does zero work

dFFddFW x

cos

Assumptions Assumptions 1) F = constant force 1) F = constant force 2) Object is particle-like (rigid object all 2) Object is particle-like (rigid object all parts of the object must move together)parts of the object must move together)

090

90180

90

W

W

A force doesA force does +W+W when it has a vector component in the when it has a vector component in the same direction as the displacement andsame direction as the displacement and ndashWndashW when it has when it has a vector component in the opposite directiona vector component in the opposite direction W=0W=0 when it when it has no such vector componenthas no such vector component

Net work done by several forces = Sum of works done by Net work done by several forces = Sum of works done by individual forcesindividual forces CalculationCalculation 1) W1) Wnetnet= W= W11+W+W22+W+W33+hellip+hellip

2) F2) Fnet net W Wnetnet=F=Fnetnet d d

II Work-Kinetic Energy TheoremII Work-Kinetic Energy Theorem

WKKK if

Change in the kinetic energy of the Change in the kinetic energy of the particle particle == Net work done on the particle Net work done on the particle

III Work done by a constant forceIII Work done by a constant force

- - Gravitational forceGravitational force

cosmgdW

Rising objectRising object W= mgd cos180W= mgd cos180ordm ordm = -mgd= -mgd FFgg transfers transfers mgdmgd energyenergy from from the objectrsquos kinetic energy the objectrsquos kinetic energy

Falling down objectFalling down object W= mgd cos 0W= mgd cos 0ordm = +mgdordm = +mgd Fg transfers transfers mgd energyenergy to to the objectrsquos kinetic energy the objectrsquos kinetic energy

External applied force + Gravitational force External applied force + Gravitational force

gaif WWKKK

Object stationary before Object stationary before and after the liftand after the lift Wa+Wg=0

The applied force transfers the same The applied force transfers the same amount of energy to the object as the amount of energy to the object as the gravitational force transfers from the objectgravitational force transfers from the object

IV Work done by a variable forceIV Work done by a variable force

- - Spring forceSpring force

xkF

Hookersquos lawHookersquos law

k = spring constant = spring constant measures springrsquos stiffness measures springrsquos stiffness Units Nm

Hookersquos LawHookersquos Law

When When xx is positive is positive (spring is stretched)(spring is stretched) FF is negativeis negative

WhenWhen xx is is 00 (at the (at the equilibrium position)equilibrium position) F F isis 00

When When xx is negative is negative (spring is compressed)(spring is compressed) FF is positiveis positive

Hookersquos LawHookersquos Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the restoring forceis called the restoring force If the block is released it will If the block is released it will

oscillate back and forth between oscillate back and forth between ndashndashxx and and xx

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

Scalar Product of Two VectorsScalar Product of Two Vectors

The scalar product The scalar product of two vectors is of two vectors is written aswritten as A A BB

It is also called It is also called the dot productthe dot product

A A bull bull B = A B cos B = A B cos θθ

θθ is the angle is the angle betweenbetween AA andand BB

Scalar ProductScalar Product

The scalar product is commutativeThe scalar product is commutative A A B = B B = B AA

The scalar product obeys the distributive law The scalar product obeys the distributive law of multiplicationof multiplication

A A (B + C) (B + C) == A A B + A B + A CC

Dot Products of Unit VectorsDot Products of Unit Vectors

bull Using component form withUsing component form with A A andand BB

0kjkiji

1kkjjii

zzyyxx

zyx

zyx

BABABABA

kBjBiBB

kAjAiAA

Work - on and byWork - on and by

A person pushes blockA person pushes block 30 m30 m along the along the ground by exerting force ofground by exerting force of 25 N25 N on the on the trolley How much work does the person do trolley How much work does the person do on the trolleyon the trolley

W = FW = Fd = 25N x 30m = 750 Nmd = 25N x 30m = 750 Nm

Trolley does -Trolley does -750 Nm750 Nm work on the personwork on the person

FFpptFFtrtr

d

A force acts on an object as A force acts on an object as the object moves in the the object moves in the xx direction from the origin direction from the origin to to xx = 500 m = 500 m Find the work Find the work W W done on the done on the object by the forceobject by the force

F 4x ˆ i 3yˆ j N

Mechanical EnergyMechanical Energy

Mechanical energy (energy associated Mechanical energy (energy associated

with with masses) can be thought of as having masses) can be thought of as having

two two components components kinetickinetic and and potentialpotential

ndashKinetic energy is energy of motionKinetic energy is energy of motion

ndashPotential energy is energy of positionPotential energy is energy of position

Kinetic EnergyKinetic Energy

bull In many situations energy can be In many situations energy can be considered asconsidered as ldquothe ability to do workrdquoldquothe ability to do workrdquo

bull Energy can exist in different formsEnergy can exist in different forms

bull Kinetic energy is the energy associated Kinetic energy is the energy associated with the with the motionmotion of an object of an object

bull A moving object can do work on A moving object can do work on another objectanother object

Eg hammer does work on nailEg hammer does work on nail

Kinetic EnergyKinetic Energy Consider an object with massConsider an object with mass mm moving in a straight line with moving in a straight line with initial velocityinitial velocity vvii To accelerate it uniformly to a speedTo accelerate it uniformly to a speed vvff a a constant net forceconstant net force FF is exerted on it parallel to motion over a is exerted on it parallel to motion over a distancedistance dd

Work done on objectWork done on object

W = F d = m a dW = F d = m a d (NII)(NII)SoSo

da

ada

if

iifif

2

2)(222

222

vv

vxxvv

dd

mW if

2

22 vv

Kinetic EnergyKinetic Energy

If we rearrange this we obtainIf we rearrange this we obtain

We define the quantityWe define the quantity frac12mvfrac12mv22 to be the to be the translational translational kinetic energykinetic energy (KE) of the object (KE) of the object

This is the This is the lsquolsquoWork-Energy TheoremWork-Energy Theoremrsquorsquo

ldquoldquoThe net The net workwork done on an object is equal to done on an object is equal to its its change in kinetic energychange in kinetic energyrdquordquo

22

2

1

2

1if mmW vv

W = KE

The Work-Energy TheoremThe Work-Energy Theorem

bull NoteNotendash The net work done on an object is the work The net work done on an object is the work

done by the net forcedone by the net forcendash Units of energy and work must be the Units of energy and work must be the

same (same (JJ))

W = W = KEKE

II Kinetic energyKinetic energy

IIII WorkWork

IIIIII Work - Kinetic energy theoremWork - Kinetic energy theorem

IVIV Work done by a constant forceWork done by a constant force

- Gravitational force- Gravitational force

V Work done by a variable forceV Work done by a variable force

- Spring force- Spring force

1D-Analysis1D-Analysis

3D-Analysis3D-Analysis

VI PowerVI Power

EnergyEnergy scalar quantity associated with a state scalar quantity associated with a state (or condition) of one or more objects (or condition) of one or more objects

II Kinetic energyKinetic energy

Energy associated with the state of motion of an objectEnergy associated with the state of motion of an object

2

2

1mvK

UnitsUnits 1 Joule = 1J = 1 kgm1 Joule = 1J = 1 kgm22ss22

II WorkII Work

Energy transferred ldquotordquo or ldquofromrdquo an Energy transferred ldquotordquo or ldquofromrdquo an object by means of a force acting on the object by means of a force acting on the object object

To To +W+WFrom From -W-W

- - Constant Constant forceforce

xx maF

dFWdFKKvvm

dvvmmaF

d

vvadavv

xxiff

fxxf

xxf

)(2

1

1)(

2

1

22

20

2

20

220

220

2

Work done by the force = Energy transfer due to the forceWork done by the force = Energy transfer due to the force

-To calculate the work done on an object by a force during To calculate the work done on an object by a force during a displacement we use only the force component alonga displacement we use only the force component along the objectrsquos displacement The force component the objectrsquos displacement The force component perpendicular to the displacement does zero workperpendicular to the displacement does zero work

dFFddFW x

cos

Assumptions Assumptions 1) F = constant force 1) F = constant force 2) Object is particle-like (rigid object all 2) Object is particle-like (rigid object all parts of the object must move together)parts of the object must move together)

090

90180

90

W

W

A force doesA force does +W+W when it has a vector component in the when it has a vector component in the same direction as the displacement andsame direction as the displacement and ndashWndashW when it has when it has a vector component in the opposite directiona vector component in the opposite direction W=0W=0 when it when it has no such vector componenthas no such vector component

Net work done by several forces = Sum of works done by Net work done by several forces = Sum of works done by individual forcesindividual forces CalculationCalculation 1) W1) Wnetnet= W= W11+W+W22+W+W33+hellip+hellip

2) F2) Fnet net W Wnetnet=F=Fnetnet d d

II Work-Kinetic Energy TheoremII Work-Kinetic Energy Theorem

WKKK if

Change in the kinetic energy of the Change in the kinetic energy of the particle particle == Net work done on the particle Net work done on the particle

III Work done by a constant forceIII Work done by a constant force

- - Gravitational forceGravitational force

cosmgdW

Rising objectRising object W= mgd cos180W= mgd cos180ordm ordm = -mgd= -mgd FFgg transfers transfers mgdmgd energyenergy from from the objectrsquos kinetic energy the objectrsquos kinetic energy

Falling down objectFalling down object W= mgd cos 0W= mgd cos 0ordm = +mgdordm = +mgd Fg transfers transfers mgd energyenergy to to the objectrsquos kinetic energy the objectrsquos kinetic energy

External applied force + Gravitational force External applied force + Gravitational force

gaif WWKKK

Object stationary before Object stationary before and after the liftand after the lift Wa+Wg=0

The applied force transfers the same The applied force transfers the same amount of energy to the object as the amount of energy to the object as the gravitational force transfers from the objectgravitational force transfers from the object

IV Work done by a variable forceIV Work done by a variable force

- - Spring forceSpring force

xkF

Hookersquos lawHookersquos law

k = spring constant = spring constant measures springrsquos stiffness measures springrsquos stiffness Units Nm

Hookersquos LawHookersquos Law

When When xx is positive is positive (spring is stretched)(spring is stretched) FF is negativeis negative

WhenWhen xx is is 00 (at the (at the equilibrium position)equilibrium position) F F isis 00

When When xx is negative is negative (spring is compressed)(spring is compressed) FF is positiveis positive

Hookersquos LawHookersquos Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the restoring forceis called the restoring force If the block is released it will If the block is released it will

oscillate back and forth between oscillate back and forth between ndashndashxx and and xx

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

Scalar ProductScalar Product

The scalar product is commutativeThe scalar product is commutative A A B = B B = B AA

The scalar product obeys the distributive law The scalar product obeys the distributive law of multiplicationof multiplication

A A (B + C) (B + C) == A A B + A B + A CC

Dot Products of Unit VectorsDot Products of Unit Vectors

bull Using component form withUsing component form with A A andand BB

0kjkiji

1kkjjii

zzyyxx

zyx

zyx

BABABABA

kBjBiBB

kAjAiAA

Work - on and byWork - on and by

A person pushes blockA person pushes block 30 m30 m along the along the ground by exerting force ofground by exerting force of 25 N25 N on the on the trolley How much work does the person do trolley How much work does the person do on the trolleyon the trolley

W = FW = Fd = 25N x 30m = 750 Nmd = 25N x 30m = 750 Nm

Trolley does -Trolley does -750 Nm750 Nm work on the personwork on the person

FFpptFFtrtr

d

A force acts on an object as A force acts on an object as the object moves in the the object moves in the xx direction from the origin direction from the origin to to xx = 500 m = 500 m Find the work Find the work W W done on the done on the object by the forceobject by the force

F 4x ˆ i 3yˆ j N

Mechanical EnergyMechanical Energy

Mechanical energy (energy associated Mechanical energy (energy associated

with with masses) can be thought of as having masses) can be thought of as having

two two components components kinetickinetic and and potentialpotential

ndashKinetic energy is energy of motionKinetic energy is energy of motion

ndashPotential energy is energy of positionPotential energy is energy of position

Kinetic EnergyKinetic Energy

bull In many situations energy can be In many situations energy can be considered asconsidered as ldquothe ability to do workrdquoldquothe ability to do workrdquo

bull Energy can exist in different formsEnergy can exist in different forms

bull Kinetic energy is the energy associated Kinetic energy is the energy associated with the with the motionmotion of an object of an object

bull A moving object can do work on A moving object can do work on another objectanother object

Eg hammer does work on nailEg hammer does work on nail

Kinetic EnergyKinetic Energy Consider an object with massConsider an object with mass mm moving in a straight line with moving in a straight line with initial velocityinitial velocity vvii To accelerate it uniformly to a speedTo accelerate it uniformly to a speed vvff a a constant net forceconstant net force FF is exerted on it parallel to motion over a is exerted on it parallel to motion over a distancedistance dd

Work done on objectWork done on object

W = F d = m a dW = F d = m a d (NII)(NII)SoSo

da

ada

if

iifif

2

2)(222

222

vv

vxxvv

dd

mW if

2

22 vv

Kinetic EnergyKinetic Energy

If we rearrange this we obtainIf we rearrange this we obtain

We define the quantityWe define the quantity frac12mvfrac12mv22 to be the to be the translational translational kinetic energykinetic energy (KE) of the object (KE) of the object

This is the This is the lsquolsquoWork-Energy TheoremWork-Energy Theoremrsquorsquo

ldquoldquoThe net The net workwork done on an object is equal to done on an object is equal to its its change in kinetic energychange in kinetic energyrdquordquo

22

2

1

2

1if mmW vv

W = KE

The Work-Energy TheoremThe Work-Energy Theorem

bull NoteNotendash The net work done on an object is the work The net work done on an object is the work

done by the net forcedone by the net forcendash Units of energy and work must be the Units of energy and work must be the

same (same (JJ))

W = W = KEKE

II Kinetic energyKinetic energy

IIII WorkWork

IIIIII Work - Kinetic energy theoremWork - Kinetic energy theorem

IVIV Work done by a constant forceWork done by a constant force

- Gravitational force- Gravitational force

V Work done by a variable forceV Work done by a variable force

- Spring force- Spring force

1D-Analysis1D-Analysis

3D-Analysis3D-Analysis

VI PowerVI Power

EnergyEnergy scalar quantity associated with a state scalar quantity associated with a state (or condition) of one or more objects (or condition) of one or more objects

II Kinetic energyKinetic energy

Energy associated with the state of motion of an objectEnergy associated with the state of motion of an object

2

2

1mvK

UnitsUnits 1 Joule = 1J = 1 kgm1 Joule = 1J = 1 kgm22ss22

II WorkII Work

Energy transferred ldquotordquo or ldquofromrdquo an Energy transferred ldquotordquo or ldquofromrdquo an object by means of a force acting on the object by means of a force acting on the object object

To To +W+WFrom From -W-W

- - Constant Constant forceforce

xx maF

dFWdFKKvvm

dvvmmaF

d

vvadavv

xxiff

fxxf

xxf

)(2

1

1)(

2

1

22

20

2

20

220

220

2

Work done by the force = Energy transfer due to the forceWork done by the force = Energy transfer due to the force

-To calculate the work done on an object by a force during To calculate the work done on an object by a force during a displacement we use only the force component alonga displacement we use only the force component along the objectrsquos displacement The force component the objectrsquos displacement The force component perpendicular to the displacement does zero workperpendicular to the displacement does zero work

dFFddFW x

cos

Assumptions Assumptions 1) F = constant force 1) F = constant force 2) Object is particle-like (rigid object all 2) Object is particle-like (rigid object all parts of the object must move together)parts of the object must move together)

090

90180

90

W

W

A force doesA force does +W+W when it has a vector component in the when it has a vector component in the same direction as the displacement andsame direction as the displacement and ndashWndashW when it has when it has a vector component in the opposite directiona vector component in the opposite direction W=0W=0 when it when it has no such vector componenthas no such vector component

Net work done by several forces = Sum of works done by Net work done by several forces = Sum of works done by individual forcesindividual forces CalculationCalculation 1) W1) Wnetnet= W= W11+W+W22+W+W33+hellip+hellip

2) F2) Fnet net W Wnetnet=F=Fnetnet d d

II Work-Kinetic Energy TheoremII Work-Kinetic Energy Theorem

WKKK if

Change in the kinetic energy of the Change in the kinetic energy of the particle particle == Net work done on the particle Net work done on the particle

III Work done by a constant forceIII Work done by a constant force

- - Gravitational forceGravitational force

cosmgdW

Rising objectRising object W= mgd cos180W= mgd cos180ordm ordm = -mgd= -mgd FFgg transfers transfers mgdmgd energyenergy from from the objectrsquos kinetic energy the objectrsquos kinetic energy

Falling down objectFalling down object W= mgd cos 0W= mgd cos 0ordm = +mgdordm = +mgd Fg transfers transfers mgd energyenergy to to the objectrsquos kinetic energy the objectrsquos kinetic energy

External applied force + Gravitational force External applied force + Gravitational force

gaif WWKKK

Object stationary before Object stationary before and after the liftand after the lift Wa+Wg=0

The applied force transfers the same The applied force transfers the same amount of energy to the object as the amount of energy to the object as the gravitational force transfers from the objectgravitational force transfers from the object

IV Work done by a variable forceIV Work done by a variable force

- - Spring forceSpring force

xkF

Hookersquos lawHookersquos law

k = spring constant = spring constant measures springrsquos stiffness measures springrsquos stiffness Units Nm

Hookersquos LawHookersquos Law

When When xx is positive is positive (spring is stretched)(spring is stretched) FF is negativeis negative

WhenWhen xx is is 00 (at the (at the equilibrium position)equilibrium position) F F isis 00

When When xx is negative is negative (spring is compressed)(spring is compressed) FF is positiveis positive

Hookersquos LawHookersquos Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the restoring forceis called the restoring force If the block is released it will If the block is released it will

oscillate back and forth between oscillate back and forth between ndashndashxx and and xx

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

Dot Products of Unit VectorsDot Products of Unit Vectors

bull Using component form withUsing component form with A A andand BB

0kjkiji

1kkjjii

zzyyxx

zyx

zyx

BABABABA

kBjBiBB

kAjAiAA

Work - on and byWork - on and by

A person pushes blockA person pushes block 30 m30 m along the along the ground by exerting force ofground by exerting force of 25 N25 N on the on the trolley How much work does the person do trolley How much work does the person do on the trolleyon the trolley

W = FW = Fd = 25N x 30m = 750 Nmd = 25N x 30m = 750 Nm

Trolley does -Trolley does -750 Nm750 Nm work on the personwork on the person

FFpptFFtrtr

d

A force acts on an object as A force acts on an object as the object moves in the the object moves in the xx direction from the origin direction from the origin to to xx = 500 m = 500 m Find the work Find the work W W done on the done on the object by the forceobject by the force

F 4x ˆ i 3yˆ j N

Mechanical EnergyMechanical Energy

Mechanical energy (energy associated Mechanical energy (energy associated

with with masses) can be thought of as having masses) can be thought of as having

two two components components kinetickinetic and and potentialpotential

ndashKinetic energy is energy of motionKinetic energy is energy of motion

ndashPotential energy is energy of positionPotential energy is energy of position

Kinetic EnergyKinetic Energy

bull In many situations energy can be In many situations energy can be considered asconsidered as ldquothe ability to do workrdquoldquothe ability to do workrdquo

bull Energy can exist in different formsEnergy can exist in different forms

bull Kinetic energy is the energy associated Kinetic energy is the energy associated with the with the motionmotion of an object of an object

bull A moving object can do work on A moving object can do work on another objectanother object

Eg hammer does work on nailEg hammer does work on nail

Kinetic EnergyKinetic Energy Consider an object with massConsider an object with mass mm moving in a straight line with moving in a straight line with initial velocityinitial velocity vvii To accelerate it uniformly to a speedTo accelerate it uniformly to a speed vvff a a constant net forceconstant net force FF is exerted on it parallel to motion over a is exerted on it parallel to motion over a distancedistance dd

Work done on objectWork done on object

W = F d = m a dW = F d = m a d (NII)(NII)SoSo

da

ada

if

iifif

2

2)(222

222

vv

vxxvv

dd

mW if

2

22 vv

Kinetic EnergyKinetic Energy

If we rearrange this we obtainIf we rearrange this we obtain

We define the quantityWe define the quantity frac12mvfrac12mv22 to be the to be the translational translational kinetic energykinetic energy (KE) of the object (KE) of the object

This is the This is the lsquolsquoWork-Energy TheoremWork-Energy Theoremrsquorsquo

ldquoldquoThe net The net workwork done on an object is equal to done on an object is equal to its its change in kinetic energychange in kinetic energyrdquordquo

22

2

1

2

1if mmW vv

W = KE

The Work-Energy TheoremThe Work-Energy Theorem

bull NoteNotendash The net work done on an object is the work The net work done on an object is the work

done by the net forcedone by the net forcendash Units of energy and work must be the Units of energy and work must be the

same (same (JJ))

W = W = KEKE

II Kinetic energyKinetic energy

IIII WorkWork

IIIIII Work - Kinetic energy theoremWork - Kinetic energy theorem

IVIV Work done by a constant forceWork done by a constant force

- Gravitational force- Gravitational force

V Work done by a variable forceV Work done by a variable force

- Spring force- Spring force

1D-Analysis1D-Analysis

3D-Analysis3D-Analysis

VI PowerVI Power

EnergyEnergy scalar quantity associated with a state scalar quantity associated with a state (or condition) of one or more objects (or condition) of one or more objects

II Kinetic energyKinetic energy

Energy associated with the state of motion of an objectEnergy associated with the state of motion of an object

2

2

1mvK

UnitsUnits 1 Joule = 1J = 1 kgm1 Joule = 1J = 1 kgm22ss22

II WorkII Work

Energy transferred ldquotordquo or ldquofromrdquo an Energy transferred ldquotordquo or ldquofromrdquo an object by means of a force acting on the object by means of a force acting on the object object

To To +W+WFrom From -W-W

- - Constant Constant forceforce

xx maF

dFWdFKKvvm

dvvmmaF

d

vvadavv

xxiff

fxxf

xxf

)(2

1

1)(

2

1

22

20

2

20

220

220

2

Work done by the force = Energy transfer due to the forceWork done by the force = Energy transfer due to the force

-To calculate the work done on an object by a force during To calculate the work done on an object by a force during a displacement we use only the force component alonga displacement we use only the force component along the objectrsquos displacement The force component the objectrsquos displacement The force component perpendicular to the displacement does zero workperpendicular to the displacement does zero work

dFFddFW x

cos

Assumptions Assumptions 1) F = constant force 1) F = constant force 2) Object is particle-like (rigid object all 2) Object is particle-like (rigid object all parts of the object must move together)parts of the object must move together)

090

90180

90

W

W

A force doesA force does +W+W when it has a vector component in the when it has a vector component in the same direction as the displacement andsame direction as the displacement and ndashWndashW when it has when it has a vector component in the opposite directiona vector component in the opposite direction W=0W=0 when it when it has no such vector componenthas no such vector component

Net work done by several forces = Sum of works done by Net work done by several forces = Sum of works done by individual forcesindividual forces CalculationCalculation 1) W1) Wnetnet= W= W11+W+W22+W+W33+hellip+hellip

2) F2) Fnet net W Wnetnet=F=Fnetnet d d

II Work-Kinetic Energy TheoremII Work-Kinetic Energy Theorem

WKKK if

Change in the kinetic energy of the Change in the kinetic energy of the particle particle == Net work done on the particle Net work done on the particle

III Work done by a constant forceIII Work done by a constant force

- - Gravitational forceGravitational force

cosmgdW

Rising objectRising object W= mgd cos180W= mgd cos180ordm ordm = -mgd= -mgd FFgg transfers transfers mgdmgd energyenergy from from the objectrsquos kinetic energy the objectrsquos kinetic energy

Falling down objectFalling down object W= mgd cos 0W= mgd cos 0ordm = +mgdordm = +mgd Fg transfers transfers mgd energyenergy to to the objectrsquos kinetic energy the objectrsquos kinetic energy

External applied force + Gravitational force External applied force + Gravitational force

gaif WWKKK

Object stationary before Object stationary before and after the liftand after the lift Wa+Wg=0

The applied force transfers the same The applied force transfers the same amount of energy to the object as the amount of energy to the object as the gravitational force transfers from the objectgravitational force transfers from the object

IV Work done by a variable forceIV Work done by a variable force

- - Spring forceSpring force

xkF

Hookersquos lawHookersquos law

k = spring constant = spring constant measures springrsquos stiffness measures springrsquos stiffness Units Nm

Hookersquos LawHookersquos Law

When When xx is positive is positive (spring is stretched)(spring is stretched) FF is negativeis negative

WhenWhen xx is is 00 (at the (at the equilibrium position)equilibrium position) F F isis 00

When When xx is negative is negative (spring is compressed)(spring is compressed) FF is positiveis positive

Hookersquos LawHookersquos Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the restoring forceis called the restoring force If the block is released it will If the block is released it will

oscillate back and forth between oscillate back and forth between ndashndashxx and and xx

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

Work - on and byWork - on and by

A person pushes blockA person pushes block 30 m30 m along the along the ground by exerting force ofground by exerting force of 25 N25 N on the on the trolley How much work does the person do trolley How much work does the person do on the trolleyon the trolley

W = FW = Fd = 25N x 30m = 750 Nmd = 25N x 30m = 750 Nm

Trolley does -Trolley does -750 Nm750 Nm work on the personwork on the person

FFpptFFtrtr

d

A force acts on an object as A force acts on an object as the object moves in the the object moves in the xx direction from the origin direction from the origin to to xx = 500 m = 500 m Find the work Find the work W W done on the done on the object by the forceobject by the force

F 4x ˆ i 3yˆ j N

Mechanical EnergyMechanical Energy

Mechanical energy (energy associated Mechanical energy (energy associated

with with masses) can be thought of as having masses) can be thought of as having

two two components components kinetickinetic and and potentialpotential

ndashKinetic energy is energy of motionKinetic energy is energy of motion

ndashPotential energy is energy of positionPotential energy is energy of position

Kinetic EnergyKinetic Energy

bull In many situations energy can be In many situations energy can be considered asconsidered as ldquothe ability to do workrdquoldquothe ability to do workrdquo

bull Energy can exist in different formsEnergy can exist in different forms

bull Kinetic energy is the energy associated Kinetic energy is the energy associated with the with the motionmotion of an object of an object

bull A moving object can do work on A moving object can do work on another objectanother object

Eg hammer does work on nailEg hammer does work on nail

Kinetic EnergyKinetic Energy Consider an object with massConsider an object with mass mm moving in a straight line with moving in a straight line with initial velocityinitial velocity vvii To accelerate it uniformly to a speedTo accelerate it uniformly to a speed vvff a a constant net forceconstant net force FF is exerted on it parallel to motion over a is exerted on it parallel to motion over a distancedistance dd

Work done on objectWork done on object

W = F d = m a dW = F d = m a d (NII)(NII)SoSo

da

ada

if

iifif

2

2)(222

222

vv

vxxvv

dd

mW if

2

22 vv

Kinetic EnergyKinetic Energy

If we rearrange this we obtainIf we rearrange this we obtain

We define the quantityWe define the quantity frac12mvfrac12mv22 to be the to be the translational translational kinetic energykinetic energy (KE) of the object (KE) of the object

This is the This is the lsquolsquoWork-Energy TheoremWork-Energy Theoremrsquorsquo

ldquoldquoThe net The net workwork done on an object is equal to done on an object is equal to its its change in kinetic energychange in kinetic energyrdquordquo

22

2

1

2

1if mmW vv

W = KE

The Work-Energy TheoremThe Work-Energy Theorem

bull NoteNotendash The net work done on an object is the work The net work done on an object is the work

done by the net forcedone by the net forcendash Units of energy and work must be the Units of energy and work must be the

same (same (JJ))

W = W = KEKE

II Kinetic energyKinetic energy

IIII WorkWork

IIIIII Work - Kinetic energy theoremWork - Kinetic energy theorem

IVIV Work done by a constant forceWork done by a constant force

- Gravitational force- Gravitational force

V Work done by a variable forceV Work done by a variable force

- Spring force- Spring force

1D-Analysis1D-Analysis

3D-Analysis3D-Analysis

VI PowerVI Power

EnergyEnergy scalar quantity associated with a state scalar quantity associated with a state (or condition) of one or more objects (or condition) of one or more objects

II Kinetic energyKinetic energy

Energy associated with the state of motion of an objectEnergy associated with the state of motion of an object

2

2

1mvK

UnitsUnits 1 Joule = 1J = 1 kgm1 Joule = 1J = 1 kgm22ss22

II WorkII Work

Energy transferred ldquotordquo or ldquofromrdquo an Energy transferred ldquotordquo or ldquofromrdquo an object by means of a force acting on the object by means of a force acting on the object object

To To +W+WFrom From -W-W

- - Constant Constant forceforce

xx maF

dFWdFKKvvm

dvvmmaF

d

vvadavv

xxiff

fxxf

xxf

)(2

1

1)(

2

1

22

20

2

20

220

220

2

Work done by the force = Energy transfer due to the forceWork done by the force = Energy transfer due to the force

-To calculate the work done on an object by a force during To calculate the work done on an object by a force during a displacement we use only the force component alonga displacement we use only the force component along the objectrsquos displacement The force component the objectrsquos displacement The force component perpendicular to the displacement does zero workperpendicular to the displacement does zero work

dFFddFW x

cos

Assumptions Assumptions 1) F = constant force 1) F = constant force 2) Object is particle-like (rigid object all 2) Object is particle-like (rigid object all parts of the object must move together)parts of the object must move together)

090

90180

90

W

W

A force doesA force does +W+W when it has a vector component in the when it has a vector component in the same direction as the displacement andsame direction as the displacement and ndashWndashW when it has when it has a vector component in the opposite directiona vector component in the opposite direction W=0W=0 when it when it has no such vector componenthas no such vector component

Net work done by several forces = Sum of works done by Net work done by several forces = Sum of works done by individual forcesindividual forces CalculationCalculation 1) W1) Wnetnet= W= W11+W+W22+W+W33+hellip+hellip

2) F2) Fnet net W Wnetnet=F=Fnetnet d d

II Work-Kinetic Energy TheoremII Work-Kinetic Energy Theorem

WKKK if

Change in the kinetic energy of the Change in the kinetic energy of the particle particle == Net work done on the particle Net work done on the particle

III Work done by a constant forceIII Work done by a constant force

- - Gravitational forceGravitational force

cosmgdW

Rising objectRising object W= mgd cos180W= mgd cos180ordm ordm = -mgd= -mgd FFgg transfers transfers mgdmgd energyenergy from from the objectrsquos kinetic energy the objectrsquos kinetic energy

Falling down objectFalling down object W= mgd cos 0W= mgd cos 0ordm = +mgdordm = +mgd Fg transfers transfers mgd energyenergy to to the objectrsquos kinetic energy the objectrsquos kinetic energy

External applied force + Gravitational force External applied force + Gravitational force

gaif WWKKK

Object stationary before Object stationary before and after the liftand after the lift Wa+Wg=0

The applied force transfers the same The applied force transfers the same amount of energy to the object as the amount of energy to the object as the gravitational force transfers from the objectgravitational force transfers from the object

IV Work done by a variable forceIV Work done by a variable force

- - Spring forceSpring force

xkF

Hookersquos lawHookersquos law

k = spring constant = spring constant measures springrsquos stiffness measures springrsquos stiffness Units Nm

Hookersquos LawHookersquos Law

When When xx is positive is positive (spring is stretched)(spring is stretched) FF is negativeis negative

WhenWhen xx is is 00 (at the (at the equilibrium position)equilibrium position) F F isis 00

When When xx is negative is negative (spring is compressed)(spring is compressed) FF is positiveis positive

Hookersquos LawHookersquos Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the restoring forceis called the restoring force If the block is released it will If the block is released it will

oscillate back and forth between oscillate back and forth between ndashndashxx and and xx

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

A force acts on an object as A force acts on an object as the object moves in the the object moves in the xx direction from the origin direction from the origin to to xx = 500 m = 500 m Find the work Find the work W W done on the done on the object by the forceobject by the force

F 4x ˆ i 3yˆ j N

Mechanical EnergyMechanical Energy

Mechanical energy (energy associated Mechanical energy (energy associated

with with masses) can be thought of as having masses) can be thought of as having

two two components components kinetickinetic and and potentialpotential

ndashKinetic energy is energy of motionKinetic energy is energy of motion

ndashPotential energy is energy of positionPotential energy is energy of position

Kinetic EnergyKinetic Energy

bull In many situations energy can be In many situations energy can be considered asconsidered as ldquothe ability to do workrdquoldquothe ability to do workrdquo

bull Energy can exist in different formsEnergy can exist in different forms

bull Kinetic energy is the energy associated Kinetic energy is the energy associated with the with the motionmotion of an object of an object

bull A moving object can do work on A moving object can do work on another objectanother object

Eg hammer does work on nailEg hammer does work on nail

Kinetic EnergyKinetic Energy Consider an object with massConsider an object with mass mm moving in a straight line with moving in a straight line with initial velocityinitial velocity vvii To accelerate it uniformly to a speedTo accelerate it uniformly to a speed vvff a a constant net forceconstant net force FF is exerted on it parallel to motion over a is exerted on it parallel to motion over a distancedistance dd

Work done on objectWork done on object

W = F d = m a dW = F d = m a d (NII)(NII)SoSo

da

ada

if

iifif

2

2)(222

222

vv

vxxvv

dd

mW if

2

22 vv

Kinetic EnergyKinetic Energy

If we rearrange this we obtainIf we rearrange this we obtain

We define the quantityWe define the quantity frac12mvfrac12mv22 to be the to be the translational translational kinetic energykinetic energy (KE) of the object (KE) of the object

This is the This is the lsquolsquoWork-Energy TheoremWork-Energy Theoremrsquorsquo

ldquoldquoThe net The net workwork done on an object is equal to done on an object is equal to its its change in kinetic energychange in kinetic energyrdquordquo

22

2

1

2

1if mmW vv

W = KE

The Work-Energy TheoremThe Work-Energy Theorem

bull NoteNotendash The net work done on an object is the work The net work done on an object is the work

done by the net forcedone by the net forcendash Units of energy and work must be the Units of energy and work must be the

same (same (JJ))

W = W = KEKE

II Kinetic energyKinetic energy

IIII WorkWork

IIIIII Work - Kinetic energy theoremWork - Kinetic energy theorem

IVIV Work done by a constant forceWork done by a constant force

- Gravitational force- Gravitational force

V Work done by a variable forceV Work done by a variable force

- Spring force- Spring force

1D-Analysis1D-Analysis

3D-Analysis3D-Analysis

VI PowerVI Power

EnergyEnergy scalar quantity associated with a state scalar quantity associated with a state (or condition) of one or more objects (or condition) of one or more objects

II Kinetic energyKinetic energy

Energy associated with the state of motion of an objectEnergy associated with the state of motion of an object

2

2

1mvK

UnitsUnits 1 Joule = 1J = 1 kgm1 Joule = 1J = 1 kgm22ss22

II WorkII Work

Energy transferred ldquotordquo or ldquofromrdquo an Energy transferred ldquotordquo or ldquofromrdquo an object by means of a force acting on the object by means of a force acting on the object object

To To +W+WFrom From -W-W

- - Constant Constant forceforce

xx maF

dFWdFKKvvm

dvvmmaF

d

vvadavv

xxiff

fxxf

xxf

)(2

1

1)(

2

1

22

20

2

20

220

220

2

Work done by the force = Energy transfer due to the forceWork done by the force = Energy transfer due to the force

-To calculate the work done on an object by a force during To calculate the work done on an object by a force during a displacement we use only the force component alonga displacement we use only the force component along the objectrsquos displacement The force component the objectrsquos displacement The force component perpendicular to the displacement does zero workperpendicular to the displacement does zero work

dFFddFW x

cos

Assumptions Assumptions 1) F = constant force 1) F = constant force 2) Object is particle-like (rigid object all 2) Object is particle-like (rigid object all parts of the object must move together)parts of the object must move together)

090

90180

90

W

W

A force doesA force does +W+W when it has a vector component in the when it has a vector component in the same direction as the displacement andsame direction as the displacement and ndashWndashW when it has when it has a vector component in the opposite directiona vector component in the opposite direction W=0W=0 when it when it has no such vector componenthas no such vector component

Net work done by several forces = Sum of works done by Net work done by several forces = Sum of works done by individual forcesindividual forces CalculationCalculation 1) W1) Wnetnet= W= W11+W+W22+W+W33+hellip+hellip

2) F2) Fnet net W Wnetnet=F=Fnetnet d d

II Work-Kinetic Energy TheoremII Work-Kinetic Energy Theorem

WKKK if

Change in the kinetic energy of the Change in the kinetic energy of the particle particle == Net work done on the particle Net work done on the particle

III Work done by a constant forceIII Work done by a constant force

- - Gravitational forceGravitational force

cosmgdW

Rising objectRising object W= mgd cos180W= mgd cos180ordm ordm = -mgd= -mgd FFgg transfers transfers mgdmgd energyenergy from from the objectrsquos kinetic energy the objectrsquos kinetic energy

Falling down objectFalling down object W= mgd cos 0W= mgd cos 0ordm = +mgdordm = +mgd Fg transfers transfers mgd energyenergy to to the objectrsquos kinetic energy the objectrsquos kinetic energy

External applied force + Gravitational force External applied force + Gravitational force

gaif WWKKK

Object stationary before Object stationary before and after the liftand after the lift Wa+Wg=0

The applied force transfers the same The applied force transfers the same amount of energy to the object as the amount of energy to the object as the gravitational force transfers from the objectgravitational force transfers from the object

IV Work done by a variable forceIV Work done by a variable force

- - Spring forceSpring force

xkF

Hookersquos lawHookersquos law

k = spring constant = spring constant measures springrsquos stiffness measures springrsquos stiffness Units Nm

Hookersquos LawHookersquos Law

When When xx is positive is positive (spring is stretched)(spring is stretched) FF is negativeis negative

WhenWhen xx is is 00 (at the (at the equilibrium position)equilibrium position) F F isis 00

When When xx is negative is negative (spring is compressed)(spring is compressed) FF is positiveis positive

Hookersquos LawHookersquos Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the restoring forceis called the restoring force If the block is released it will If the block is released it will

oscillate back and forth between oscillate back and forth between ndashndashxx and and xx

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

Mechanical EnergyMechanical Energy

Mechanical energy (energy associated Mechanical energy (energy associated

with with masses) can be thought of as having masses) can be thought of as having

two two components components kinetickinetic and and potentialpotential

ndashKinetic energy is energy of motionKinetic energy is energy of motion

ndashPotential energy is energy of positionPotential energy is energy of position

Kinetic EnergyKinetic Energy

bull In many situations energy can be In many situations energy can be considered asconsidered as ldquothe ability to do workrdquoldquothe ability to do workrdquo

bull Energy can exist in different formsEnergy can exist in different forms

bull Kinetic energy is the energy associated Kinetic energy is the energy associated with the with the motionmotion of an object of an object

bull A moving object can do work on A moving object can do work on another objectanother object

Eg hammer does work on nailEg hammer does work on nail

Kinetic EnergyKinetic Energy Consider an object with massConsider an object with mass mm moving in a straight line with moving in a straight line with initial velocityinitial velocity vvii To accelerate it uniformly to a speedTo accelerate it uniformly to a speed vvff a a constant net forceconstant net force FF is exerted on it parallel to motion over a is exerted on it parallel to motion over a distancedistance dd

Work done on objectWork done on object

W = F d = m a dW = F d = m a d (NII)(NII)SoSo

da

ada

if

iifif

2

2)(222

222

vv

vxxvv

dd

mW if

2

22 vv

Kinetic EnergyKinetic Energy

If we rearrange this we obtainIf we rearrange this we obtain

We define the quantityWe define the quantity frac12mvfrac12mv22 to be the to be the translational translational kinetic energykinetic energy (KE) of the object (KE) of the object

This is the This is the lsquolsquoWork-Energy TheoremWork-Energy Theoremrsquorsquo

ldquoldquoThe net The net workwork done on an object is equal to done on an object is equal to its its change in kinetic energychange in kinetic energyrdquordquo

22

2

1

2

1if mmW vv

W = KE

The Work-Energy TheoremThe Work-Energy Theorem

bull NoteNotendash The net work done on an object is the work The net work done on an object is the work

done by the net forcedone by the net forcendash Units of energy and work must be the Units of energy and work must be the

same (same (JJ))

W = W = KEKE

II Kinetic energyKinetic energy

IIII WorkWork

IIIIII Work - Kinetic energy theoremWork - Kinetic energy theorem

IVIV Work done by a constant forceWork done by a constant force

- Gravitational force- Gravitational force

V Work done by a variable forceV Work done by a variable force

- Spring force- Spring force

1D-Analysis1D-Analysis

3D-Analysis3D-Analysis

VI PowerVI Power

EnergyEnergy scalar quantity associated with a state scalar quantity associated with a state (or condition) of one or more objects (or condition) of one or more objects

II Kinetic energyKinetic energy

Energy associated with the state of motion of an objectEnergy associated with the state of motion of an object

2

2

1mvK

UnitsUnits 1 Joule = 1J = 1 kgm1 Joule = 1J = 1 kgm22ss22

II WorkII Work

Energy transferred ldquotordquo or ldquofromrdquo an Energy transferred ldquotordquo or ldquofromrdquo an object by means of a force acting on the object by means of a force acting on the object object

To To +W+WFrom From -W-W

- - Constant Constant forceforce

xx maF

dFWdFKKvvm

dvvmmaF

d

vvadavv

xxiff

fxxf

xxf

)(2

1

1)(

2

1

22

20

2

20

220

220

2

Work done by the force = Energy transfer due to the forceWork done by the force = Energy transfer due to the force

-To calculate the work done on an object by a force during To calculate the work done on an object by a force during a displacement we use only the force component alonga displacement we use only the force component along the objectrsquos displacement The force component the objectrsquos displacement The force component perpendicular to the displacement does zero workperpendicular to the displacement does zero work

dFFddFW x

cos

Assumptions Assumptions 1) F = constant force 1) F = constant force 2) Object is particle-like (rigid object all 2) Object is particle-like (rigid object all parts of the object must move together)parts of the object must move together)

090

90180

90

W

W

A force doesA force does +W+W when it has a vector component in the when it has a vector component in the same direction as the displacement andsame direction as the displacement and ndashWndashW when it has when it has a vector component in the opposite directiona vector component in the opposite direction W=0W=0 when it when it has no such vector componenthas no such vector component

Net work done by several forces = Sum of works done by Net work done by several forces = Sum of works done by individual forcesindividual forces CalculationCalculation 1) W1) Wnetnet= W= W11+W+W22+W+W33+hellip+hellip

2) F2) Fnet net W Wnetnet=F=Fnetnet d d

II Work-Kinetic Energy TheoremII Work-Kinetic Energy Theorem

WKKK if

Change in the kinetic energy of the Change in the kinetic energy of the particle particle == Net work done on the particle Net work done on the particle

III Work done by a constant forceIII Work done by a constant force

- - Gravitational forceGravitational force

cosmgdW

Rising objectRising object W= mgd cos180W= mgd cos180ordm ordm = -mgd= -mgd FFgg transfers transfers mgdmgd energyenergy from from the objectrsquos kinetic energy the objectrsquos kinetic energy

Falling down objectFalling down object W= mgd cos 0W= mgd cos 0ordm = +mgdordm = +mgd Fg transfers transfers mgd energyenergy to to the objectrsquos kinetic energy the objectrsquos kinetic energy

External applied force + Gravitational force External applied force + Gravitational force

gaif WWKKK

Object stationary before Object stationary before and after the liftand after the lift Wa+Wg=0

The applied force transfers the same The applied force transfers the same amount of energy to the object as the amount of energy to the object as the gravitational force transfers from the objectgravitational force transfers from the object

IV Work done by a variable forceIV Work done by a variable force

- - Spring forceSpring force

xkF

Hookersquos lawHookersquos law

k = spring constant = spring constant measures springrsquos stiffness measures springrsquos stiffness Units Nm

Hookersquos LawHookersquos Law

When When xx is positive is positive (spring is stretched)(spring is stretched) FF is negativeis negative

WhenWhen xx is is 00 (at the (at the equilibrium position)equilibrium position) F F isis 00

When When xx is negative is negative (spring is compressed)(spring is compressed) FF is positiveis positive

Hookersquos LawHookersquos Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the restoring forceis called the restoring force If the block is released it will If the block is released it will

oscillate back and forth between oscillate back and forth between ndashndashxx and and xx

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

Kinetic EnergyKinetic Energy

bull In many situations energy can be In many situations energy can be considered asconsidered as ldquothe ability to do workrdquoldquothe ability to do workrdquo

bull Energy can exist in different formsEnergy can exist in different forms

bull Kinetic energy is the energy associated Kinetic energy is the energy associated with the with the motionmotion of an object of an object

bull A moving object can do work on A moving object can do work on another objectanother object

Eg hammer does work on nailEg hammer does work on nail

Kinetic EnergyKinetic Energy Consider an object with massConsider an object with mass mm moving in a straight line with moving in a straight line with initial velocityinitial velocity vvii To accelerate it uniformly to a speedTo accelerate it uniformly to a speed vvff a a constant net forceconstant net force FF is exerted on it parallel to motion over a is exerted on it parallel to motion over a distancedistance dd

Work done on objectWork done on object

W = F d = m a dW = F d = m a d (NII)(NII)SoSo

da

ada

if

iifif

2

2)(222

222

vv

vxxvv

dd

mW if

2

22 vv

Kinetic EnergyKinetic Energy

If we rearrange this we obtainIf we rearrange this we obtain

We define the quantityWe define the quantity frac12mvfrac12mv22 to be the to be the translational translational kinetic energykinetic energy (KE) of the object (KE) of the object

This is the This is the lsquolsquoWork-Energy TheoremWork-Energy Theoremrsquorsquo

ldquoldquoThe net The net workwork done on an object is equal to done on an object is equal to its its change in kinetic energychange in kinetic energyrdquordquo

22

2

1

2

1if mmW vv

W = KE

The Work-Energy TheoremThe Work-Energy Theorem

bull NoteNotendash The net work done on an object is the work The net work done on an object is the work

done by the net forcedone by the net forcendash Units of energy and work must be the Units of energy and work must be the

same (same (JJ))

W = W = KEKE

II Kinetic energyKinetic energy

IIII WorkWork

IIIIII Work - Kinetic energy theoremWork - Kinetic energy theorem

IVIV Work done by a constant forceWork done by a constant force

- Gravitational force- Gravitational force

V Work done by a variable forceV Work done by a variable force

- Spring force- Spring force

1D-Analysis1D-Analysis

3D-Analysis3D-Analysis

VI PowerVI Power

EnergyEnergy scalar quantity associated with a state scalar quantity associated with a state (or condition) of one or more objects (or condition) of one or more objects

II Kinetic energyKinetic energy

Energy associated with the state of motion of an objectEnergy associated with the state of motion of an object

2

2

1mvK

UnitsUnits 1 Joule = 1J = 1 kgm1 Joule = 1J = 1 kgm22ss22

II WorkII Work

Energy transferred ldquotordquo or ldquofromrdquo an Energy transferred ldquotordquo or ldquofromrdquo an object by means of a force acting on the object by means of a force acting on the object object

To To +W+WFrom From -W-W

- - Constant Constant forceforce

xx maF

dFWdFKKvvm

dvvmmaF

d

vvadavv

xxiff

fxxf

xxf

)(2

1

1)(

2

1

22

20

2

20

220

220

2

Work done by the force = Energy transfer due to the forceWork done by the force = Energy transfer due to the force

-To calculate the work done on an object by a force during To calculate the work done on an object by a force during a displacement we use only the force component alonga displacement we use only the force component along the objectrsquos displacement The force component the objectrsquos displacement The force component perpendicular to the displacement does zero workperpendicular to the displacement does zero work

dFFddFW x

cos

Assumptions Assumptions 1) F = constant force 1) F = constant force 2) Object is particle-like (rigid object all 2) Object is particle-like (rigid object all parts of the object must move together)parts of the object must move together)

090

90180

90

W

W

A force doesA force does +W+W when it has a vector component in the when it has a vector component in the same direction as the displacement andsame direction as the displacement and ndashWndashW when it has when it has a vector component in the opposite directiona vector component in the opposite direction W=0W=0 when it when it has no such vector componenthas no such vector component

Net work done by several forces = Sum of works done by Net work done by several forces = Sum of works done by individual forcesindividual forces CalculationCalculation 1) W1) Wnetnet= W= W11+W+W22+W+W33+hellip+hellip

2) F2) Fnet net W Wnetnet=F=Fnetnet d d

II Work-Kinetic Energy TheoremII Work-Kinetic Energy Theorem

WKKK if

Change in the kinetic energy of the Change in the kinetic energy of the particle particle == Net work done on the particle Net work done on the particle

III Work done by a constant forceIII Work done by a constant force

- - Gravitational forceGravitational force

cosmgdW

Rising objectRising object W= mgd cos180W= mgd cos180ordm ordm = -mgd= -mgd FFgg transfers transfers mgdmgd energyenergy from from the objectrsquos kinetic energy the objectrsquos kinetic energy

Falling down objectFalling down object W= mgd cos 0W= mgd cos 0ordm = +mgdordm = +mgd Fg transfers transfers mgd energyenergy to to the objectrsquos kinetic energy the objectrsquos kinetic energy

External applied force + Gravitational force External applied force + Gravitational force

gaif WWKKK

Object stationary before Object stationary before and after the liftand after the lift Wa+Wg=0

The applied force transfers the same The applied force transfers the same amount of energy to the object as the amount of energy to the object as the gravitational force transfers from the objectgravitational force transfers from the object

IV Work done by a variable forceIV Work done by a variable force

- - Spring forceSpring force

xkF

Hookersquos lawHookersquos law

k = spring constant = spring constant measures springrsquos stiffness measures springrsquos stiffness Units Nm

Hookersquos LawHookersquos Law

When When xx is positive is positive (spring is stretched)(spring is stretched) FF is negativeis negative

WhenWhen xx is is 00 (at the (at the equilibrium position)equilibrium position) F F isis 00

When When xx is negative is negative (spring is compressed)(spring is compressed) FF is positiveis positive

Hookersquos LawHookersquos Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the restoring forceis called the restoring force If the block is released it will If the block is released it will

oscillate back and forth between oscillate back and forth between ndashndashxx and and xx

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

Kinetic EnergyKinetic Energy Consider an object with massConsider an object with mass mm moving in a straight line with moving in a straight line with initial velocityinitial velocity vvii To accelerate it uniformly to a speedTo accelerate it uniformly to a speed vvff a a constant net forceconstant net force FF is exerted on it parallel to motion over a is exerted on it parallel to motion over a distancedistance dd

Work done on objectWork done on object

W = F d = m a dW = F d = m a d (NII)(NII)SoSo

da

ada

if

iifif

2

2)(222

222

vv

vxxvv

dd

mW if

2

22 vv

Kinetic EnergyKinetic Energy

If we rearrange this we obtainIf we rearrange this we obtain

We define the quantityWe define the quantity frac12mvfrac12mv22 to be the to be the translational translational kinetic energykinetic energy (KE) of the object (KE) of the object

This is the This is the lsquolsquoWork-Energy TheoremWork-Energy Theoremrsquorsquo

ldquoldquoThe net The net workwork done on an object is equal to done on an object is equal to its its change in kinetic energychange in kinetic energyrdquordquo

22

2

1

2

1if mmW vv

W = KE

The Work-Energy TheoremThe Work-Energy Theorem

bull NoteNotendash The net work done on an object is the work The net work done on an object is the work

done by the net forcedone by the net forcendash Units of energy and work must be the Units of energy and work must be the

same (same (JJ))

W = W = KEKE

II Kinetic energyKinetic energy

IIII WorkWork

IIIIII Work - Kinetic energy theoremWork - Kinetic energy theorem

IVIV Work done by a constant forceWork done by a constant force

- Gravitational force- Gravitational force

V Work done by a variable forceV Work done by a variable force

- Spring force- Spring force

1D-Analysis1D-Analysis

3D-Analysis3D-Analysis

VI PowerVI Power

EnergyEnergy scalar quantity associated with a state scalar quantity associated with a state (or condition) of one or more objects (or condition) of one or more objects

II Kinetic energyKinetic energy

Energy associated with the state of motion of an objectEnergy associated with the state of motion of an object

2

2

1mvK

UnitsUnits 1 Joule = 1J = 1 kgm1 Joule = 1J = 1 kgm22ss22

II WorkII Work

Energy transferred ldquotordquo or ldquofromrdquo an Energy transferred ldquotordquo or ldquofromrdquo an object by means of a force acting on the object by means of a force acting on the object object

To To +W+WFrom From -W-W

- - Constant Constant forceforce

xx maF

dFWdFKKvvm

dvvmmaF

d

vvadavv

xxiff

fxxf

xxf

)(2

1

1)(

2

1

22

20

2

20

220

220

2

Work done by the force = Energy transfer due to the forceWork done by the force = Energy transfer due to the force

-To calculate the work done on an object by a force during To calculate the work done on an object by a force during a displacement we use only the force component alonga displacement we use only the force component along the objectrsquos displacement The force component the objectrsquos displacement The force component perpendicular to the displacement does zero workperpendicular to the displacement does zero work

dFFddFW x

cos

Assumptions Assumptions 1) F = constant force 1) F = constant force 2) Object is particle-like (rigid object all 2) Object is particle-like (rigid object all parts of the object must move together)parts of the object must move together)

090

90180

90

W

W

A force doesA force does +W+W when it has a vector component in the when it has a vector component in the same direction as the displacement andsame direction as the displacement and ndashWndashW when it has when it has a vector component in the opposite directiona vector component in the opposite direction W=0W=0 when it when it has no such vector componenthas no such vector component

Net work done by several forces = Sum of works done by Net work done by several forces = Sum of works done by individual forcesindividual forces CalculationCalculation 1) W1) Wnetnet= W= W11+W+W22+W+W33+hellip+hellip

2) F2) Fnet net W Wnetnet=F=Fnetnet d d

II Work-Kinetic Energy TheoremII Work-Kinetic Energy Theorem

WKKK if

Change in the kinetic energy of the Change in the kinetic energy of the particle particle == Net work done on the particle Net work done on the particle

III Work done by a constant forceIII Work done by a constant force

- - Gravitational forceGravitational force

cosmgdW

Rising objectRising object W= mgd cos180W= mgd cos180ordm ordm = -mgd= -mgd FFgg transfers transfers mgdmgd energyenergy from from the objectrsquos kinetic energy the objectrsquos kinetic energy

Falling down objectFalling down object W= mgd cos 0W= mgd cos 0ordm = +mgdordm = +mgd Fg transfers transfers mgd energyenergy to to the objectrsquos kinetic energy the objectrsquos kinetic energy

External applied force + Gravitational force External applied force + Gravitational force

gaif WWKKK

Object stationary before Object stationary before and after the liftand after the lift Wa+Wg=0

The applied force transfers the same The applied force transfers the same amount of energy to the object as the amount of energy to the object as the gravitational force transfers from the objectgravitational force transfers from the object

IV Work done by a variable forceIV Work done by a variable force

- - Spring forceSpring force

xkF

Hookersquos lawHookersquos law

k = spring constant = spring constant measures springrsquos stiffness measures springrsquos stiffness Units Nm

Hookersquos LawHookersquos Law

When When xx is positive is positive (spring is stretched)(spring is stretched) FF is negativeis negative

WhenWhen xx is is 00 (at the (at the equilibrium position)equilibrium position) F F isis 00

When When xx is negative is negative (spring is compressed)(spring is compressed) FF is positiveis positive

Hookersquos LawHookersquos Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the restoring forceis called the restoring force If the block is released it will If the block is released it will

oscillate back and forth between oscillate back and forth between ndashndashxx and and xx

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

Kinetic EnergyKinetic Energy

If we rearrange this we obtainIf we rearrange this we obtain

We define the quantityWe define the quantity frac12mvfrac12mv22 to be the to be the translational translational kinetic energykinetic energy (KE) of the object (KE) of the object

This is the This is the lsquolsquoWork-Energy TheoremWork-Energy Theoremrsquorsquo

ldquoldquoThe net The net workwork done on an object is equal to done on an object is equal to its its change in kinetic energychange in kinetic energyrdquordquo

22

2

1

2

1if mmW vv

W = KE

The Work-Energy TheoremThe Work-Energy Theorem

bull NoteNotendash The net work done on an object is the work The net work done on an object is the work

done by the net forcedone by the net forcendash Units of energy and work must be the Units of energy and work must be the

same (same (JJ))

W = W = KEKE

II Kinetic energyKinetic energy

IIII WorkWork

IIIIII Work - Kinetic energy theoremWork - Kinetic energy theorem

IVIV Work done by a constant forceWork done by a constant force

- Gravitational force- Gravitational force

V Work done by a variable forceV Work done by a variable force

- Spring force- Spring force

1D-Analysis1D-Analysis

3D-Analysis3D-Analysis

VI PowerVI Power

EnergyEnergy scalar quantity associated with a state scalar quantity associated with a state (or condition) of one or more objects (or condition) of one or more objects

II Kinetic energyKinetic energy

Energy associated with the state of motion of an objectEnergy associated with the state of motion of an object

2

2

1mvK

UnitsUnits 1 Joule = 1J = 1 kgm1 Joule = 1J = 1 kgm22ss22

II WorkII Work

Energy transferred ldquotordquo or ldquofromrdquo an Energy transferred ldquotordquo or ldquofromrdquo an object by means of a force acting on the object by means of a force acting on the object object

To To +W+WFrom From -W-W

- - Constant Constant forceforce

xx maF

dFWdFKKvvm

dvvmmaF

d

vvadavv

xxiff

fxxf

xxf

)(2

1

1)(

2

1

22

20

2

20

220

220

2

Work done by the force = Energy transfer due to the forceWork done by the force = Energy transfer due to the force

-To calculate the work done on an object by a force during To calculate the work done on an object by a force during a displacement we use only the force component alonga displacement we use only the force component along the objectrsquos displacement The force component the objectrsquos displacement The force component perpendicular to the displacement does zero workperpendicular to the displacement does zero work

dFFddFW x

cos

Assumptions Assumptions 1) F = constant force 1) F = constant force 2) Object is particle-like (rigid object all 2) Object is particle-like (rigid object all parts of the object must move together)parts of the object must move together)

090

90180

90

W

W

A force doesA force does +W+W when it has a vector component in the when it has a vector component in the same direction as the displacement andsame direction as the displacement and ndashWndashW when it has when it has a vector component in the opposite directiona vector component in the opposite direction W=0W=0 when it when it has no such vector componenthas no such vector component

Net work done by several forces = Sum of works done by Net work done by several forces = Sum of works done by individual forcesindividual forces CalculationCalculation 1) W1) Wnetnet= W= W11+W+W22+W+W33+hellip+hellip

2) F2) Fnet net W Wnetnet=F=Fnetnet d d

II Work-Kinetic Energy TheoremII Work-Kinetic Energy Theorem

WKKK if

Change in the kinetic energy of the Change in the kinetic energy of the particle particle == Net work done on the particle Net work done on the particle

III Work done by a constant forceIII Work done by a constant force

- - Gravitational forceGravitational force

cosmgdW

Rising objectRising object W= mgd cos180W= mgd cos180ordm ordm = -mgd= -mgd FFgg transfers transfers mgdmgd energyenergy from from the objectrsquos kinetic energy the objectrsquos kinetic energy

Falling down objectFalling down object W= mgd cos 0W= mgd cos 0ordm = +mgdordm = +mgd Fg transfers transfers mgd energyenergy to to the objectrsquos kinetic energy the objectrsquos kinetic energy

External applied force + Gravitational force External applied force + Gravitational force

gaif WWKKK

Object stationary before Object stationary before and after the liftand after the lift Wa+Wg=0

The applied force transfers the same The applied force transfers the same amount of energy to the object as the amount of energy to the object as the gravitational force transfers from the objectgravitational force transfers from the object

IV Work done by a variable forceIV Work done by a variable force

- - Spring forceSpring force

xkF

Hookersquos lawHookersquos law

k = spring constant = spring constant measures springrsquos stiffness measures springrsquos stiffness Units Nm

Hookersquos LawHookersquos Law

When When xx is positive is positive (spring is stretched)(spring is stretched) FF is negativeis negative

WhenWhen xx is is 00 (at the (at the equilibrium position)equilibrium position) F F isis 00

When When xx is negative is negative (spring is compressed)(spring is compressed) FF is positiveis positive

Hookersquos LawHookersquos Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the restoring forceis called the restoring force If the block is released it will If the block is released it will

oscillate back and forth between oscillate back and forth between ndashndashxx and and xx

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

The Work-Energy TheoremThe Work-Energy Theorem

bull NoteNotendash The net work done on an object is the work The net work done on an object is the work

done by the net forcedone by the net forcendash Units of energy and work must be the Units of energy and work must be the

same (same (JJ))

W = W = KEKE

II Kinetic energyKinetic energy

IIII WorkWork

IIIIII Work - Kinetic energy theoremWork - Kinetic energy theorem

IVIV Work done by a constant forceWork done by a constant force

- Gravitational force- Gravitational force

V Work done by a variable forceV Work done by a variable force

- Spring force- Spring force

1D-Analysis1D-Analysis

3D-Analysis3D-Analysis

VI PowerVI Power

EnergyEnergy scalar quantity associated with a state scalar quantity associated with a state (or condition) of one or more objects (or condition) of one or more objects

II Kinetic energyKinetic energy

Energy associated with the state of motion of an objectEnergy associated with the state of motion of an object

2

2

1mvK

UnitsUnits 1 Joule = 1J = 1 kgm1 Joule = 1J = 1 kgm22ss22

II WorkII Work

Energy transferred ldquotordquo or ldquofromrdquo an Energy transferred ldquotordquo or ldquofromrdquo an object by means of a force acting on the object by means of a force acting on the object object

To To +W+WFrom From -W-W

- - Constant Constant forceforce

xx maF

dFWdFKKvvm

dvvmmaF

d

vvadavv

xxiff

fxxf

xxf

)(2

1

1)(

2

1

22

20

2

20

220

220

2

Work done by the force = Energy transfer due to the forceWork done by the force = Energy transfer due to the force

-To calculate the work done on an object by a force during To calculate the work done on an object by a force during a displacement we use only the force component alonga displacement we use only the force component along the objectrsquos displacement The force component the objectrsquos displacement The force component perpendicular to the displacement does zero workperpendicular to the displacement does zero work

dFFddFW x

cos

Assumptions Assumptions 1) F = constant force 1) F = constant force 2) Object is particle-like (rigid object all 2) Object is particle-like (rigid object all parts of the object must move together)parts of the object must move together)

090

90180

90

W

W

A force doesA force does +W+W when it has a vector component in the when it has a vector component in the same direction as the displacement andsame direction as the displacement and ndashWndashW when it has when it has a vector component in the opposite directiona vector component in the opposite direction W=0W=0 when it when it has no such vector componenthas no such vector component

Net work done by several forces = Sum of works done by Net work done by several forces = Sum of works done by individual forcesindividual forces CalculationCalculation 1) W1) Wnetnet= W= W11+W+W22+W+W33+hellip+hellip

2) F2) Fnet net W Wnetnet=F=Fnetnet d d

II Work-Kinetic Energy TheoremII Work-Kinetic Energy Theorem

WKKK if

Change in the kinetic energy of the Change in the kinetic energy of the particle particle == Net work done on the particle Net work done on the particle

III Work done by a constant forceIII Work done by a constant force

- - Gravitational forceGravitational force

cosmgdW

Rising objectRising object W= mgd cos180W= mgd cos180ordm ordm = -mgd= -mgd FFgg transfers transfers mgdmgd energyenergy from from the objectrsquos kinetic energy the objectrsquos kinetic energy

Falling down objectFalling down object W= mgd cos 0W= mgd cos 0ordm = +mgdordm = +mgd Fg transfers transfers mgd energyenergy to to the objectrsquos kinetic energy the objectrsquos kinetic energy

External applied force + Gravitational force External applied force + Gravitational force

gaif WWKKK

Object stationary before Object stationary before and after the liftand after the lift Wa+Wg=0

The applied force transfers the same The applied force transfers the same amount of energy to the object as the amount of energy to the object as the gravitational force transfers from the objectgravitational force transfers from the object

IV Work done by a variable forceIV Work done by a variable force

- - Spring forceSpring force

xkF

Hookersquos lawHookersquos law

k = spring constant = spring constant measures springrsquos stiffness measures springrsquos stiffness Units Nm

Hookersquos LawHookersquos Law

When When xx is positive is positive (spring is stretched)(spring is stretched) FF is negativeis negative

WhenWhen xx is is 00 (at the (at the equilibrium position)equilibrium position) F F isis 00

When When xx is negative is negative (spring is compressed)(spring is compressed) FF is positiveis positive

Hookersquos LawHookersquos Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the restoring forceis called the restoring force If the block is released it will If the block is released it will

oscillate back and forth between oscillate back and forth between ndashndashxx and and xx

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

II Kinetic energyKinetic energy

IIII WorkWork

IIIIII Work - Kinetic energy theoremWork - Kinetic energy theorem

IVIV Work done by a constant forceWork done by a constant force

- Gravitational force- Gravitational force

V Work done by a variable forceV Work done by a variable force

- Spring force- Spring force

1D-Analysis1D-Analysis

3D-Analysis3D-Analysis

VI PowerVI Power

EnergyEnergy scalar quantity associated with a state scalar quantity associated with a state (or condition) of one or more objects (or condition) of one or more objects

II Kinetic energyKinetic energy

Energy associated with the state of motion of an objectEnergy associated with the state of motion of an object

2

2

1mvK

UnitsUnits 1 Joule = 1J = 1 kgm1 Joule = 1J = 1 kgm22ss22

II WorkII Work

Energy transferred ldquotordquo or ldquofromrdquo an Energy transferred ldquotordquo or ldquofromrdquo an object by means of a force acting on the object by means of a force acting on the object object

To To +W+WFrom From -W-W

- - Constant Constant forceforce

xx maF

dFWdFKKvvm

dvvmmaF

d

vvadavv

xxiff

fxxf

xxf

)(2

1

1)(

2

1

22

20

2

20

220

220

2

Work done by the force = Energy transfer due to the forceWork done by the force = Energy transfer due to the force

-To calculate the work done on an object by a force during To calculate the work done on an object by a force during a displacement we use only the force component alonga displacement we use only the force component along the objectrsquos displacement The force component the objectrsquos displacement The force component perpendicular to the displacement does zero workperpendicular to the displacement does zero work

dFFddFW x

cos

Assumptions Assumptions 1) F = constant force 1) F = constant force 2) Object is particle-like (rigid object all 2) Object is particle-like (rigid object all parts of the object must move together)parts of the object must move together)

090

90180

90

W

W

A force doesA force does +W+W when it has a vector component in the when it has a vector component in the same direction as the displacement andsame direction as the displacement and ndashWndashW when it has when it has a vector component in the opposite directiona vector component in the opposite direction W=0W=0 when it when it has no such vector componenthas no such vector component

Net work done by several forces = Sum of works done by Net work done by several forces = Sum of works done by individual forcesindividual forces CalculationCalculation 1) W1) Wnetnet= W= W11+W+W22+W+W33+hellip+hellip

2) F2) Fnet net W Wnetnet=F=Fnetnet d d

II Work-Kinetic Energy TheoremII Work-Kinetic Energy Theorem

WKKK if

Change in the kinetic energy of the Change in the kinetic energy of the particle particle == Net work done on the particle Net work done on the particle

III Work done by a constant forceIII Work done by a constant force

- - Gravitational forceGravitational force

cosmgdW

Rising objectRising object W= mgd cos180W= mgd cos180ordm ordm = -mgd= -mgd FFgg transfers transfers mgdmgd energyenergy from from the objectrsquos kinetic energy the objectrsquos kinetic energy

Falling down objectFalling down object W= mgd cos 0W= mgd cos 0ordm = +mgdordm = +mgd Fg transfers transfers mgd energyenergy to to the objectrsquos kinetic energy the objectrsquos kinetic energy

External applied force + Gravitational force External applied force + Gravitational force

gaif WWKKK

Object stationary before Object stationary before and after the liftand after the lift Wa+Wg=0

The applied force transfers the same The applied force transfers the same amount of energy to the object as the amount of energy to the object as the gravitational force transfers from the objectgravitational force transfers from the object

IV Work done by a variable forceIV Work done by a variable force

- - Spring forceSpring force

xkF

Hookersquos lawHookersquos law

k = spring constant = spring constant measures springrsquos stiffness measures springrsquos stiffness Units Nm

Hookersquos LawHookersquos Law

When When xx is positive is positive (spring is stretched)(spring is stretched) FF is negativeis negative

WhenWhen xx is is 00 (at the (at the equilibrium position)equilibrium position) F F isis 00

When When xx is negative is negative (spring is compressed)(spring is compressed) FF is positiveis positive

Hookersquos LawHookersquos Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the restoring forceis called the restoring force If the block is released it will If the block is released it will

oscillate back and forth between oscillate back and forth between ndashndashxx and and xx

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

II Kinetic energyKinetic energy

Energy associated with the state of motion of an objectEnergy associated with the state of motion of an object

2

2

1mvK

UnitsUnits 1 Joule = 1J = 1 kgm1 Joule = 1J = 1 kgm22ss22

II WorkII Work

Energy transferred ldquotordquo or ldquofromrdquo an Energy transferred ldquotordquo or ldquofromrdquo an object by means of a force acting on the object by means of a force acting on the object object

To To +W+WFrom From -W-W

- - Constant Constant forceforce

xx maF

dFWdFKKvvm

dvvmmaF

d

vvadavv

xxiff

fxxf

xxf

)(2

1

1)(

2

1

22

20

2

20

220

220

2

Work done by the force = Energy transfer due to the forceWork done by the force = Energy transfer due to the force

-To calculate the work done on an object by a force during To calculate the work done on an object by a force during a displacement we use only the force component alonga displacement we use only the force component along the objectrsquos displacement The force component the objectrsquos displacement The force component perpendicular to the displacement does zero workperpendicular to the displacement does zero work

dFFddFW x

cos

Assumptions Assumptions 1) F = constant force 1) F = constant force 2) Object is particle-like (rigid object all 2) Object is particle-like (rigid object all parts of the object must move together)parts of the object must move together)

090

90180

90

W

W

A force doesA force does +W+W when it has a vector component in the when it has a vector component in the same direction as the displacement andsame direction as the displacement and ndashWndashW when it has when it has a vector component in the opposite directiona vector component in the opposite direction W=0W=0 when it when it has no such vector componenthas no such vector component

Net work done by several forces = Sum of works done by Net work done by several forces = Sum of works done by individual forcesindividual forces CalculationCalculation 1) W1) Wnetnet= W= W11+W+W22+W+W33+hellip+hellip

2) F2) Fnet net W Wnetnet=F=Fnetnet d d

II Work-Kinetic Energy TheoremII Work-Kinetic Energy Theorem

WKKK if

Change in the kinetic energy of the Change in the kinetic energy of the particle particle == Net work done on the particle Net work done on the particle

III Work done by a constant forceIII Work done by a constant force

- - Gravitational forceGravitational force

cosmgdW

Rising objectRising object W= mgd cos180W= mgd cos180ordm ordm = -mgd= -mgd FFgg transfers transfers mgdmgd energyenergy from from the objectrsquos kinetic energy the objectrsquos kinetic energy

Falling down objectFalling down object W= mgd cos 0W= mgd cos 0ordm = +mgdordm = +mgd Fg transfers transfers mgd energyenergy to to the objectrsquos kinetic energy the objectrsquos kinetic energy

External applied force + Gravitational force External applied force + Gravitational force

gaif WWKKK

Object stationary before Object stationary before and after the liftand after the lift Wa+Wg=0

The applied force transfers the same The applied force transfers the same amount of energy to the object as the amount of energy to the object as the gravitational force transfers from the objectgravitational force transfers from the object

IV Work done by a variable forceIV Work done by a variable force

- - Spring forceSpring force

xkF

Hookersquos lawHookersquos law

k = spring constant = spring constant measures springrsquos stiffness measures springrsquos stiffness Units Nm

Hookersquos LawHookersquos Law

When When xx is positive is positive (spring is stretched)(spring is stretched) FF is negativeis negative

WhenWhen xx is is 00 (at the (at the equilibrium position)equilibrium position) F F isis 00

When When xx is negative is negative (spring is compressed)(spring is compressed) FF is positiveis positive

Hookersquos LawHookersquos Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the restoring forceis called the restoring force If the block is released it will If the block is released it will

oscillate back and forth between oscillate back and forth between ndashndashxx and and xx

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

- - Constant Constant forceforce

xx maF

dFWdFKKvvm

dvvmmaF

d

vvadavv

xxiff

fxxf

xxf

)(2

1

1)(

2

1

22

20

2

20

220

220

2

Work done by the force = Energy transfer due to the forceWork done by the force = Energy transfer due to the force

-To calculate the work done on an object by a force during To calculate the work done on an object by a force during a displacement we use only the force component alonga displacement we use only the force component along the objectrsquos displacement The force component the objectrsquos displacement The force component perpendicular to the displacement does zero workperpendicular to the displacement does zero work

dFFddFW x

cos

Assumptions Assumptions 1) F = constant force 1) F = constant force 2) Object is particle-like (rigid object all 2) Object is particle-like (rigid object all parts of the object must move together)parts of the object must move together)

090

90180

90

W

W

A force doesA force does +W+W when it has a vector component in the when it has a vector component in the same direction as the displacement andsame direction as the displacement and ndashWndashW when it has when it has a vector component in the opposite directiona vector component in the opposite direction W=0W=0 when it when it has no such vector componenthas no such vector component

Net work done by several forces = Sum of works done by Net work done by several forces = Sum of works done by individual forcesindividual forces CalculationCalculation 1) W1) Wnetnet= W= W11+W+W22+W+W33+hellip+hellip

2) F2) Fnet net W Wnetnet=F=Fnetnet d d

II Work-Kinetic Energy TheoremII Work-Kinetic Energy Theorem

WKKK if

Change in the kinetic energy of the Change in the kinetic energy of the particle particle == Net work done on the particle Net work done on the particle

III Work done by a constant forceIII Work done by a constant force

- - Gravitational forceGravitational force

cosmgdW

Rising objectRising object W= mgd cos180W= mgd cos180ordm ordm = -mgd= -mgd FFgg transfers transfers mgdmgd energyenergy from from the objectrsquos kinetic energy the objectrsquos kinetic energy

Falling down objectFalling down object W= mgd cos 0W= mgd cos 0ordm = +mgdordm = +mgd Fg transfers transfers mgd energyenergy to to the objectrsquos kinetic energy the objectrsquos kinetic energy

External applied force + Gravitational force External applied force + Gravitational force

gaif WWKKK

Object stationary before Object stationary before and after the liftand after the lift Wa+Wg=0

The applied force transfers the same The applied force transfers the same amount of energy to the object as the amount of energy to the object as the gravitational force transfers from the objectgravitational force transfers from the object

IV Work done by a variable forceIV Work done by a variable force

- - Spring forceSpring force

xkF

Hookersquos lawHookersquos law

k = spring constant = spring constant measures springrsquos stiffness measures springrsquos stiffness Units Nm

Hookersquos LawHookersquos Law

When When xx is positive is positive (spring is stretched)(spring is stretched) FF is negativeis negative

WhenWhen xx is is 00 (at the (at the equilibrium position)equilibrium position) F F isis 00

When When xx is negative is negative (spring is compressed)(spring is compressed) FF is positiveis positive

Hookersquos LawHookersquos Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the restoring forceis called the restoring force If the block is released it will If the block is released it will

oscillate back and forth between oscillate back and forth between ndashndashxx and and xx

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

-To calculate the work done on an object by a force during To calculate the work done on an object by a force during a displacement we use only the force component alonga displacement we use only the force component along the objectrsquos displacement The force component the objectrsquos displacement The force component perpendicular to the displacement does zero workperpendicular to the displacement does zero work

dFFddFW x

cos

Assumptions Assumptions 1) F = constant force 1) F = constant force 2) Object is particle-like (rigid object all 2) Object is particle-like (rigid object all parts of the object must move together)parts of the object must move together)

090

90180

90

W

W

A force doesA force does +W+W when it has a vector component in the when it has a vector component in the same direction as the displacement andsame direction as the displacement and ndashWndashW when it has when it has a vector component in the opposite directiona vector component in the opposite direction W=0W=0 when it when it has no such vector componenthas no such vector component

Net work done by several forces = Sum of works done by Net work done by several forces = Sum of works done by individual forcesindividual forces CalculationCalculation 1) W1) Wnetnet= W= W11+W+W22+W+W33+hellip+hellip

2) F2) Fnet net W Wnetnet=F=Fnetnet d d

II Work-Kinetic Energy TheoremII Work-Kinetic Energy Theorem

WKKK if

Change in the kinetic energy of the Change in the kinetic energy of the particle particle == Net work done on the particle Net work done on the particle

III Work done by a constant forceIII Work done by a constant force

- - Gravitational forceGravitational force

cosmgdW

Rising objectRising object W= mgd cos180W= mgd cos180ordm ordm = -mgd= -mgd FFgg transfers transfers mgdmgd energyenergy from from the objectrsquos kinetic energy the objectrsquos kinetic energy

Falling down objectFalling down object W= mgd cos 0W= mgd cos 0ordm = +mgdordm = +mgd Fg transfers transfers mgd energyenergy to to the objectrsquos kinetic energy the objectrsquos kinetic energy

External applied force + Gravitational force External applied force + Gravitational force

gaif WWKKK

Object stationary before Object stationary before and after the liftand after the lift Wa+Wg=0

The applied force transfers the same The applied force transfers the same amount of energy to the object as the amount of energy to the object as the gravitational force transfers from the objectgravitational force transfers from the object

IV Work done by a variable forceIV Work done by a variable force

- - Spring forceSpring force

xkF

Hookersquos lawHookersquos law

k = spring constant = spring constant measures springrsquos stiffness measures springrsquos stiffness Units Nm

Hookersquos LawHookersquos Law

When When xx is positive is positive (spring is stretched)(spring is stretched) FF is negativeis negative

WhenWhen xx is is 00 (at the (at the equilibrium position)equilibrium position) F F isis 00

When When xx is negative is negative (spring is compressed)(spring is compressed) FF is positiveis positive

Hookersquos LawHookersquos Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the restoring forceis called the restoring force If the block is released it will If the block is released it will

oscillate back and forth between oscillate back and forth between ndashndashxx and and xx

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

Net work done by several forces = Sum of works done by Net work done by several forces = Sum of works done by individual forcesindividual forces CalculationCalculation 1) W1) Wnetnet= W= W11+W+W22+W+W33+hellip+hellip

2) F2) Fnet net W Wnetnet=F=Fnetnet d d

II Work-Kinetic Energy TheoremII Work-Kinetic Energy Theorem

WKKK if

Change in the kinetic energy of the Change in the kinetic energy of the particle particle == Net work done on the particle Net work done on the particle

III Work done by a constant forceIII Work done by a constant force

- - Gravitational forceGravitational force

cosmgdW

Rising objectRising object W= mgd cos180W= mgd cos180ordm ordm = -mgd= -mgd FFgg transfers transfers mgdmgd energyenergy from from the objectrsquos kinetic energy the objectrsquos kinetic energy

Falling down objectFalling down object W= mgd cos 0W= mgd cos 0ordm = +mgdordm = +mgd Fg transfers transfers mgd energyenergy to to the objectrsquos kinetic energy the objectrsquos kinetic energy

External applied force + Gravitational force External applied force + Gravitational force

gaif WWKKK

Object stationary before Object stationary before and after the liftand after the lift Wa+Wg=0

The applied force transfers the same The applied force transfers the same amount of energy to the object as the amount of energy to the object as the gravitational force transfers from the objectgravitational force transfers from the object

IV Work done by a variable forceIV Work done by a variable force

- - Spring forceSpring force

xkF

Hookersquos lawHookersquos law

k = spring constant = spring constant measures springrsquos stiffness measures springrsquos stiffness Units Nm

Hookersquos LawHookersquos Law

When When xx is positive is positive (spring is stretched)(spring is stretched) FF is negativeis negative

WhenWhen xx is is 00 (at the (at the equilibrium position)equilibrium position) F F isis 00

When When xx is negative is negative (spring is compressed)(spring is compressed) FF is positiveis positive

Hookersquos LawHookersquos Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the restoring forceis called the restoring force If the block is released it will If the block is released it will

oscillate back and forth between oscillate back and forth between ndashndashxx and and xx

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

II Work-Kinetic Energy TheoremII Work-Kinetic Energy Theorem

WKKK if

Change in the kinetic energy of the Change in the kinetic energy of the particle particle == Net work done on the particle Net work done on the particle

III Work done by a constant forceIII Work done by a constant force

- - Gravitational forceGravitational force

cosmgdW

Rising objectRising object W= mgd cos180W= mgd cos180ordm ordm = -mgd= -mgd FFgg transfers transfers mgdmgd energyenergy from from the objectrsquos kinetic energy the objectrsquos kinetic energy

Falling down objectFalling down object W= mgd cos 0W= mgd cos 0ordm = +mgdordm = +mgd Fg transfers transfers mgd energyenergy to to the objectrsquos kinetic energy the objectrsquos kinetic energy

External applied force + Gravitational force External applied force + Gravitational force

gaif WWKKK

Object stationary before Object stationary before and after the liftand after the lift Wa+Wg=0

The applied force transfers the same The applied force transfers the same amount of energy to the object as the amount of energy to the object as the gravitational force transfers from the objectgravitational force transfers from the object

IV Work done by a variable forceIV Work done by a variable force

- - Spring forceSpring force

xkF

Hookersquos lawHookersquos law

k = spring constant = spring constant measures springrsquos stiffness measures springrsquos stiffness Units Nm

Hookersquos LawHookersquos Law

When When xx is positive is positive (spring is stretched)(spring is stretched) FF is negativeis negative

WhenWhen xx is is 00 (at the (at the equilibrium position)equilibrium position) F F isis 00

When When xx is negative is negative (spring is compressed)(spring is compressed) FF is positiveis positive

Hookersquos LawHookersquos Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the restoring forceis called the restoring force If the block is released it will If the block is released it will

oscillate back and forth between oscillate back and forth between ndashndashxx and and xx

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

External applied force + Gravitational force External applied force + Gravitational force

gaif WWKKK

Object stationary before Object stationary before and after the liftand after the lift Wa+Wg=0

The applied force transfers the same The applied force transfers the same amount of energy to the object as the amount of energy to the object as the gravitational force transfers from the objectgravitational force transfers from the object

IV Work done by a variable forceIV Work done by a variable force

- - Spring forceSpring force

xkF

Hookersquos lawHookersquos law

k = spring constant = spring constant measures springrsquos stiffness measures springrsquos stiffness Units Nm

Hookersquos LawHookersquos Law

When When xx is positive is positive (spring is stretched)(spring is stretched) FF is negativeis negative

WhenWhen xx is is 00 (at the (at the equilibrium position)equilibrium position) F F isis 00

When When xx is negative is negative (spring is compressed)(spring is compressed) FF is positiveis positive

Hookersquos LawHookersquos Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the restoring forceis called the restoring force If the block is released it will If the block is released it will

oscillate back and forth between oscillate back and forth between ndashndashxx and and xx

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

IV Work done by a variable forceIV Work done by a variable force

- - Spring forceSpring force

xkF

Hookersquos lawHookersquos law

k = spring constant = spring constant measures springrsquos stiffness measures springrsquos stiffness Units Nm

Hookersquos LawHookersquos Law

When When xx is positive is positive (spring is stretched)(spring is stretched) FF is negativeis negative

WhenWhen xx is is 00 (at the (at the equilibrium position)equilibrium position) F F isis 00

When When xx is negative is negative (spring is compressed)(spring is compressed) FF is positiveis positive

Hookersquos LawHookersquos Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the restoring forceis called the restoring force If the block is released it will If the block is released it will

oscillate back and forth between oscillate back and forth between ndashndashxx and and xx

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

Hookersquos LawHookersquos Law

When When xx is positive is positive (spring is stretched)(spring is stretched) FF is negativeis negative

WhenWhen xx is is 00 (at the (at the equilibrium position)equilibrium position) F F isis 00

When When xx is negative is negative (spring is compressed)(spring is compressed) FF is positiveis positive

Hookersquos LawHookersquos Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the restoring forceis called the restoring force If the block is released it will If the block is released it will

oscillate back and forth between oscillate back and forth between ndashndashxx and and xx

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

Hookersquos LawHookersquos Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the restoring forceis called the restoring force If the block is released it will If the block is released it will

oscillate back and forth between oscillate back and forth between ndashndashxx and and xx

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

kxFD x 1

Work done by a spring forceWork done by a spring force

Hookersquos lawHookersquos law

AssumptionsAssumptions

bull Spring is massless Spring is massless mspring ltlt mblock

bull Ideal spring Ideal spring obeys Hookersquos law exactly obeys Hookersquos law exactlybull Contact between the block and floor is frictionlessContact between the block and floor is frictionlessbull Block is particle-likeBlock is particle-like

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal widthinto many segments of infinitesimal width Δx

2)2) F(x) asymp constant within each shortwithin each short Δx segmentsegment

- Calculation- Calculation

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWssgt0 gt0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initiallyrelaxed position (x=0) than it was initiallyWWsslt0lt0 Block ends up further away from Block ends up further away from

x=0x=0WWss=0=0 Block ends up at Block ends up at x=0x=0

02

1 2 ifs xifxkW

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

Work done by an applied force+spring Work done by an applied force+spring forceforce

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

If it takesIf it takes 400 J400 J of work to stretch a Hookes-of work to stretch a Hookes-law springlaw spring 100 cm100 cm from its unstressed from its unstressed length determine the extra work required to length determine the extra work required to stretch it an additionalstretch it an additional 100 cm100 cm

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

AA 200-kg200-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 Nm500 Nm The block is pulledThe block is pulled 500 cm500 cm to the to the right of equilibrium and released from rest Find the right of equilibrium and released from rest Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 03500350

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

Work done by a general variable forceWork done by a general variable force

bull Assume that during a Assume that during a very small displacementvery small displacement ΔΔxx FF is constantis constant

bull For that displacementFor that displacement WW ~ ~ FF ΔΔxx

bull For all of the intervalsFor all of the intervals

f

i

x

xx

W F x

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

Work done by a general variable forceWork done by a general variable force

bull

bull ThereforeTherefore

bull The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)()()(ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

V PowerV Power Time rate at which the applied force does workTime rate at which the applied force does work

- Average power- Average power amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force by a force

-Instantaneous power Instantaneous power -instantaneous time rate of doing workinstantaneous time rate of doing work

t

WPavg

dt

dWP

UnitsUnits 1 watt= 1 W = 1Js1 watt= 1 W = 1Js 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000Wmiddotmiddoth = 1000Js x 3600s = 36 x 10h = 1000Js x 3600s = 36 x 1066 J J = 36 MJ= 36 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor Find an expression for the speedacross a frictionless floor Find an expression for the speed vvff at at

the end of that distance if the blockrsquos initial velocity is (a)the end of that distance if the blockrsquos initial velocity is (a) 00 and and (b)(b) 1ms1ms to the rightto the right

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

(c) The situation in fig(b) is similar in that the block is initially (c) The situation in fig(b) is similar in that the block is initially moving atmoving at 1ms1ms to the right but now theto the right but now the 2N2N force is directed force is directed downward to the left Find an expression for the speed of the downward to the left Find an expression for the speed of the block at the end of the 1m distanceblock at the end of the 1m distance

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder as illustrated in Figure (a) If the over the top of the cylinder as illustrated in Figure (a) If the particle moves at a constant speed show that particle moves at a constant speed show that FF = = mgmgcoscos ((NoteNote If the particle moves at constant speed the If the particle moves at constant speed the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times) (b) By directly integrating be zero at all times) (b) By directly integrating W = W = FFddrr find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder from the bottom to the top of the half-cylinder

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

Two springs with negligible masses one with spring Two springs with negligible masses one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22 are are attached to the endstops of a level air track as in Figure A attached to the endstops of a level air track as in Figure A glider attached to both springs is located between them glider attached to both springs is located between them When the glider is in equilibrium spring When the glider is in equilibrium spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left Now a horizontal force to the left Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position Show that in this process (a) the its equilibrium position Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) (b) the work done (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 ndash 2 ndash 2xxaaxxii22)) (c) (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11kk22 and (d) the total work done by the force and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

N6 AA 2kg2kg lunchbox is sent sliding over a frictionless surface lunchbox is sent sliding over a frictionless surface in the positive direction of anin the positive direction of an xx axis along the surface axis along the surface Beginning atBeginning at t=0t=0 a steady wind pushes on the lunchbox in the a steady wind pushes on the lunchbox in the negative direction ofnegative direction of x x so itrsquos position is changed as so itrsquos position is changed as x=t-01tx=t-01t22 Estimate the kinetic energy of the lunchbox at (a)Estimate the kinetic energy of the lunchbox at (a) t=1s t=1s (b) (b) t=5st=5s (c) How much work does the force from the wind do on (c) How much work does the force from the wind do on the lunch box fromthe lunch box from t=1st=1s to to t=5st=5s

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

N12 In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book as the book slides a distance ofbook as the book slides a distance of d=05m d=05m up a frictionless ramp (a) During the displacement up a frictionless ramp (a) During the displacement what is the net work done on the book bywhat is the net work done on the book by FFaa the gravitational the gravitational

force on the book and the normal force on the book (b) If the force on the book and the normal force on the book (b) If the book has zero kinetic energy at the start of the displacement book has zero kinetic energy at the start of the displacement what is the speed at the end of the displacementwhat is the speed at the end of the displacement

x

y

mg

N

Fgy

Fgx

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

N15 (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m (b) The curve is given (b) The curve is givenby by F=axF=ax22 with with a=9Nma=9Nm22 Calculate the work using integration Calculate the work using integration

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

N19N19 An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg If the cab is moving upward at full load at constant If the cab is moving upward at full load at constant speed speed 38ms38ms what power is required of the force moving the cab what power is required of the force moving the cab to maintain that speedto maintain that speed

mgmg

FFaa

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

N17 A single force acts on a body that moves along an x-axis A single force acts on a body that moves along an x-axis The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body For each of the intervals the body For each of the intervals AB BC CD and DE give the signAB BC CD and DE give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force or state that the work is the force or state that the work is zerozero

vv

ttAA

BB CC

DD

EE

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

15E15E In the In the figure below a cord runs around two massless figure below a cord runs around two massless frictionless pulleys a canister with mass frictionless pulleys a canister with mass m=20kgm=20kg hangs from one hangs from one pulley and you exert a force pulley and you exert a force FF on the free end of the cord on the free end of the cord (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed (b) To lift the canister by speed (b) To lift the canister by 2cm2cm how far must you pull the free end of how far must you pull the free end of the cord During that lift what is the the cord During that lift what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canistergravitational force on the canistermg

TT T

P1

P2

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

Challenging problems ndash Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod The trolleys lie on a horizontal frictionless floor A man wishes rigid rod The trolleys lie on a horizontal frictionless floor A man wishes to push them with a force of 1200N to push them with a force of 1200N From the point of view of kinetics From the point of view of kinetics does the relative position of the does the relative position of the trolleys matter If the rod can trolleys matter If the rod can only stand an applied force of 500N only stand an applied force of 500N which trolley should be up front which trolley should be up front

Situation 1

m1g m2g

N1 N2

F

F1r F2r

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56

22 A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a constant velocity of develops a constant velocity of 112 kmhour112 kmhour when traveling when traveling along a horizontal straight highway Assuming that the along a horizontal straight highway Assuming that the frictional forces (from the ground and air) acting on the car are frictional forces (from the ground and air) acting on the car are constant but not negligible What is the value of the frictional constant but not negligible What is the value of the frictional forces forces (a) What is the carrsquos maximum velocity on a (a) What is the carrsquos maximum velocity on a 5500 incline hill (b) incline hill (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36kmh36kmh

• Energy and Work
• Introduction to Energy
• Work and Energy
• Energy Approach to Problems
• Slide 5
• Systems
• Valid System
• Environment
• Work
• Slide 10
• Work Example
• More About Work
• Work Is An Energy Transfer
• Work done by a constant force
• Work
• Scalar Product of Two Vectors
• Scalar Product
• Dot Products of Unit Vectors
• Work - on and by
• PowerPoint Presentation
• Mechanical Energy
• Kinetic Energy
• Kinetic Energy
• Slide 24
• The Work-Energy Theorem
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Hookersquos Law
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Work done by a general variable force
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56