AP Physics Enosburg Falls High School

Mr. Bushey

Week 6: Work, Energy, Power

Homework ! Read Giancoli Chapter 6.1 6.10 AND/OR Read Saxon

Lessons 12, 16, 29, 48 ! Read Topic Summary Handout ! Answer Giancoli p.174 Problems 1, 4, 7, 17, 18, 19, 20, 21, 23,

25 ! Answer Princeton Review p.79 AP Set multiple choice, free

response questions

Supplemental Review / Study Group Focus ! Review Key Concepts Handout ! Schaums Outlines, Chapter 6 Problems 6.24 6.53 ! Schaums Outlines, Chapter 7 Problems 7.10 7.17 ! Solve problems from associated Saxon lessons

Work and Work-Energy Theorem

Work

Work is defined as the scalar product of force and displacement. Work is done when a non-zero net force acts on a moving object. If a force F moves an object over a distance d, work can be

found by using the formula:

W F d!!!

"

If F and d are parallel, the work is W F d! . If they are not parallel, then the work is

cosW F d "! .

From this formula, one can conclude that for the work to be done, the following conditions must

be met:

# The object must move (i.e. d $ 0). A force can be exerted on an object with no work done,

e.g. pushing on a wall, holding up 100 lb, etc.

# If F and d are perpendicular, no work is done (cos" = 0). For example, carrying something is

not work because the angle between the force (upwards) and displacement (horizontal) is

equal to 0.

# The SI unit of work is Joule (J), and 1 J = 1 N x 1 m.

# If the angle between force and displacement is less than 900, cos" >0, and the work is

positive. If 90o

Energy

Energy is one of the most fundamental concepts of Physics. It is a very abstract concept that

does not have a single definition. Energy is a characteristic of a system. Energy in a system gives

it the capability to perform some operation. When work is done on a system or by a system, the

energy of the system changes. In other words, the system gains or loses energy through the

process of work. If positive work is done, the system gains energy. If work is negative, the system

loses energy. If no work is done, the total energy of the system does not change.

There are many types of energy. Kinetic energy (K or Ek) is the energy of motion. It depends only

upon the square of the speed and the mass of the object.

212mvK !

Work - Energy Theorem

An object is moving with an acceleration a over a distance d. Then according to the formula,

: 2 2 2f iv v a d! "

2 2 2( )f iFv v dm

! " . F!

F!

d!

vi vf

Solving for F d gives

2 2( )2 f imF d v v! #

The work done on the object W is given by

2 2 2 21 1( )2 2 2f i f i f i

mW F d v v mv mv K K! ! # ! # ! #

Or, we can rewrite it as

ifW K K K! # !$

This is the kinetic energywork theorem: The change of kinetic energy of a body is equal to

the work of all the forces acting upon it.

Work and Work-Energy Theorem

1. A box, initially at rest, is pulled a distance of 5 m across a floor by a horizontal force of 23 N. At the end of the 5 m, the kinetic energy of the box is 87 J.

a. What was the net work done on the box?

According to the Work-Energy theorem, the net work done on the box is equal to the change in the kinetic energy.

87Jnet kW E! " !

b. How much work was done by the 23 N force?

W = F d = (23 N)(5 m) = 115 J

c. How much work was done by friction? The net work is equal to the sum of the work done by the applied force and the work done by the friction.

Wnet = WA + Wf

Wf = Wnet WA =87 J 115 J = 28 J

d. How much work was done by the gravitational force? The gravitational force is perpendicular to the displacement. Thus, the gravitational force does no work. Wg = 0 J. 2. Bill carries a 35 N package from the ground up to the fifth floor of a 15 m high office building. How much work is done by Bill on the package? The force Bill is applying to the package is directed up. He must be applying a force directed up to the package that is equal to the packages weight. Hes also moving up. Therefore, # = 0. F = 35 N W = F d cos# = F d d = 15 m W = (35 N)(15 m)(cos 0) = 525 J W = ?

3. A 4 kg brick slides a distance (d) of 5m along an icy slope of inclination angle of 30o. The

coefficient of kinetic friction is !k = 0.20. What will be the speed of the brick at the end of the

slope?

"

There are three forces acting on the brick.

The work done by gravity = m g d cos(90 - "), as the angle enclosed by G and the slope is 90 - ".

Note that cos(90 - ") = sin" (trigonometry), therefore, Wg = m g d sin ". Note that d sin" = h

where h is the height of the slope. Thus, the work is equal to m g h.

The normal force encloses 90o with the displacement, so Wn = Fn d cos(90o) = 0.

The work of the friction is Wfr = - !#Fn#d = - !k m g d#cos(").

The work of all forces is W = m g d (sin " - !k cos ") and this will be equal to the kinetic energy Kf

= 1/2 mvf2.

2

2

2

(sin cos )2

2 (sin cos )

2 (sin cos )

2(9.81m/s )(5m)(0.500 (0.20)(0.866))5.7 m/s

f

f

f

mvmgd

gd v

v gd

" ! "

" ! "

" ! "

$ %

$ %

% $

% $

%

4. A 2 kg block is accelerated from rest along a horizontal, smooth surface by a force of 5 N over a distance of 6 m. a. Determine the acceleration of the block and the final velocity of the block.

25 N 2.50m/s2 kg

Fam

% % %

& ' & '& '

2 2

22 2

2 2

2

0 m/s 2 2.50m/s 6 m 30 m /s

30 m /s 5.48 m/s

f o

f

f

v v a x

v

v

$ % #(

$ % %

% %

2 2

b. Calculate the work done by the accelerating force. ! "! " ! "o 5 N 6 m Cos 0 30 N m 30.0 JW F d Cos #$ $ $ % $ c. Use the Work-Energy theorem to determine the final velocity of the block. Compare your answer to the answer obtained in part (a).

! "

i

2f

2 2f

-

30 J 0 J 30 J

1 2kg v 30 J2

30 Jv 30m /s 5.48 m/s1 kg

f

f i

W KE KE KEKE W KE

$ & $

$ ' $ ' $

$

$ $ $

d. Find the final velocity of the block if it has an initial velocity of +2 m/s.

! " ! "! "

! "

2 2

22

2

2 2

-

1 1 2 21 130J = 2 kg 2 kg 2 m/s 2 2

30J = 1 kg 4 J

34 J 34 m /s 5.83 m/s1 kg

f i

f i

f

f

f

W KE KE KE

W mv mv

v

v

v

$ & $

$ (

(

(

$ $ $

e. If the block is moving with a velocity of 8 m/s, what magnitude retarding force is needed to bring the block to rest over a distance of 5 m?

! " ! " ! "! "

2 2

2 2

-

1 1 - 2 21 1 2 kg 0 m/s - 2 kg 8 m/s2 2

-64 J

f i

f i

KE KE KE

m v m v

& $

$

$

$

-64 J (5 m)(1)

-12.8 N

KEFx Cos

#&

$& %

$

$

5. Hookes law states that F k x! " . a. Find the work required to compress the spring through a distance x. Because the force is not constant throughout the distance moved, we need to integrate the product of the distance and the force to determine the work. Also the force we exert is opposite in direction to the force exerted by the spring, so the force we exert is F k x! .

22 2

0 00

1 102 2 2

xx x xW F dx k x dx k k x k x! ! ! ! " !# #

The work required is then 212

W k x! .

b. What is the potential energy of a spring which was initially at zero and was compressed to a point x? When the spring has been compressed its velocity is zero, so there is no kinetic energy, all of the energy is then potential energy.

Work is equal to the change in the energy. Thus, 212spring

U k! x .

Conservative Forces and Potential Energy

Conservative Forces

Definition: A force is called conservative if the work it does on an object depends only on the initial and final positions of the object and is independent of the path taken between those positions.

Gravity is one such conservative force. Near Earths surface, the work done by gravity on an object of mass m depends only on the change in the objects height h.

In the case of conservative forces, we assign a number--the potential energy, U--to each configuration of the system. The zero level of the potential energy is arbitrary; it can be assigned

to any position. If the place with zero potential energy is chosen, the potential energy at any point

A is the work done by the force when the body moves from the point A to the point 0 with zero

potential energy. So, U(A) = WA0.

The work is additive. If the body moves from A to B and then to point 0, then

, 0 0A ABW W W! " B

or . ( ) ( )ABU A W U B! " ( ) ( )ABW U A U B! #

Examples of Potential Energy

The Gravitational Potential Energy

Gravity is an example of a conservative force. The work done on an object of mass m when it is

raised height h at constant velocity by an external force is mgh. Ug = mgh is called the

gravitational potential energy of the object.

The only physically significant quantity is the change in potential energy, and not the absolute

value of the potential energy. For this reason, in problem solving, we usually choose a convenient

point (table top level, ground level, sea level, etc.) and set the potential energy at this level to be

zero. All heights are measured from this level. Then, if the object is below the zero level, its

potential energy is negative. It is important to stress again that only the change in potential

energy is physically significant.

Gravitational Potential Energy: Ug = mgy

Change in Gravitational Potential Energy: $og g g

U = U - U = mgh - mgho

The Elastic Potential Energy

Springs, rubber bands, and other springy objects also exert conservative forces. At relatively

small elongations, the force exerted by a spring is proportional to the amount x by which the

spring has been stretched,

sprF k x! " ,

(The - sign indicates that the force is directed in the opposite direction to the stretch.)

k is the constant of proportionality and is called the springs spring constant. This proportionality between the force F and the amount of stretching (or compression) x is known as Hookes Law. The work done by a spring with a given spring constant depends only on x, so an associated

potential energy is

21spr 2( )U x k x! .

Note that x = 0 is at the equilibrium (unstretched) position of the spring.

Nonconservative forces

Unlike conservative forces, the work done by nonconservative forces depends on the path the

object moves and not just on the initial and final points. Friction is an example of a

nonconservative force. A nonconservative force is often called a dissipative force.

Conservative Forces and Potential Energy 1. A 5 kg bowling ball is carried to the top of a tower that is 45 m high. The bowling ball is released from rest and falling freely to the ground. a. What force is required to lift the bowling ball to the top of the tower at constant velocity?

Wt = 50 N

FApp = 50 N Net

0 N -

= (5kg)(10 m/s) 50 NApplied

Applied

F F WF W m g

! !

! ! !

b. How much work is done by the lifting force to lift the bowling ball from the ground to the top of the tower? " #" # 50 N 45 m 2250 JNetW F d! ! ! c. What is the Gravitational Potential Energy of the bowling ball on the top of the tower? 2250 JtopGPE W! ! $ d. How do the GPE and the KE of the bowling ball change if the bowling ball falls from the top of the tower to the ground? As the bowling ball falls, it loses height and its GPE decreases. Because of the gravitational force, the bowling ball accelerates (g = 10 m/s2), its speed increases, and its KE increases. e. Use the equations of motion to find the speed of the bowling ball at the moment it hits the ground. What is the KE of the bowling ball at the instant it strikes the ground? Compare it with the answer from part (c).

" #" #

" #" #

20

2 2

2 2 2

2 2

22

45 m ; 0 m/s ; 10 m/s

2

(0 m/s) 2 10 m/s 45 m 900 m /s

900 m /s 30 m/s

1 1 5 kg 30 m/s 2250 J2 2

f o

f

f

k f f

d v av v a d

v

v

E mv

! % ! ! %

% !

% ! % % !

! !

! ! !

2 2

The answer is the same as in part (c).

f. Use the equations of motion for free fall to determine the height above the ground and the potential energy of the bowling ball at t = 2 s after it is released.

! "! " ! " ! "

! " ! "! "

20 0

22

2

1 2

145m 0 m/s 2 s 10m/s 2 s2

25 m

5 kg 10m/s 25 m 1250 J

f

f

f

h h v t a t

h

GPE mgh

# $ $

# $ $ %

#

# # #

g. Calculate the velocity and the KE of the bowling ball at time t = 2 s after it is released.

! " ! "

! "

0

2

2

2

0 m/s 10 m/s 2 s 20 m/s

1 21 (5 kg) 20 m/s 1000 J2

f

f f

v v a t

KE m v

# $

# $ % # %

#

# % #

h. Calculate the sum of the KE and GPE of the bowling ball at this time and compare it with its initial potential energy and with its final kinetic energy at the moment it hits the ground. GPEf + KEf = 1000J + 1250J = 2250J = GPEtop = KEf i. The bowling ball hits the ground and compresses the ground a distance of 0.020 m. What is the magnitude of the average force exerted by the ground on the bowling ball as it comes to rest? Ignoring the small change in GPE as the ground is compressed, the Change in KE is equal to the work done by the force exerted by the ground:

0 J - 2250 J -2250 J

Work Done = ( )

( 0.020 m 0 ) 2250 J2250 J 113,000 N

0.020 m

f o

f i

KE KE KE

KEF d F d d KEF m

F

& # % # #

# &% # &

% % # %%

# #%

Conservation of Energy and Power

According to the Work-energy Theorem,

_All ForcesW K! "

Work of all forces can be separated into two components, work of conservative forces and work

of non-conservative forces.

_ _ _All Forces Consearvative Forces Non Conservative ForcesW W W! # _

K

U

Further,

_Conservative ForcesW U! $"

Then the work energy theorem becomes

_ _Non Conservative ForcesW U$" ! "

_ _Non Conservative ForcesW K! " # "

This leads to the concept of the conservation of mechanical energy.

If there is no non-conservative force in the system, _ _ 0Non Conservative ForcesW !

and 0 K U! " # " .

In the absence of dissipative forces in a mechanical system, the mechanical energy Emech= K + U

is a constant. For example, Emech in time t = tinitial is the same as Emech at time t = tfinal.

The work energy theorem gives the result

0mechE K U" ! " # " !

In an isolated system where only conservative forces cause energy to be transformed, the

mechanical energy of the system is constant.

If no non-conservative forces are acting in a system, then 0 K U! " # " .

When there are forces such as friction present, we can still include these forces and conserve

total energy taking into account the heat generated as a result of work of dissipative forces such

as friction.

For Friction, nc k kW f d N!" # " $ d!!" !"

This work takes away mechanical energy and transfers this detracted amount into heat.

From the work energy theorem,

k Nd K U!$ " % & %

0 kK U Nd!" % & % &

0 mech thermalE E" % & %

Thermal energy is one of the forms of non-mechanical internal energy stored inside of the bodies.

Internal energy will be covered in the following units of the course. The conservation of energy

with inclusion of internal energy becomes

int0 mechE E" % & %

If there are external forces applied to an isolated system as a whole then, the work energy theorem is

intext mechF d E E# " % & %!!!" !"

or,

extF Total

W E" %

All of this can by summed up in the following equation:

Total final mechanical energy

= + - Total initial

mechanical energyAny work done

Any losses due tononconservative forces

or

app lossf f i iK U K U W W& " & & $

Power

Power is the rate (how fast) at which work is done.

Work DonePower . Time Spent Performing Work

!

WPt

!

The unit of power is the "Watt," which is equal to a Joule/second. The "horsepower" is the English system unit of power; it is equal to 746 W.

Conservation of Energy and Power 1. An electrical motor lifts a 575 N box 20 m straight up by a rope in 10 s. What power is developed by the motor? F = 575 N the amount of force the rope must apply to the object to lift it up at constant velocity is equal to the objects weight

F = 575 N WPt

!

d = 20 m W F d!

t = 10 s FdPt

!

P = ? 3(575N)(20m) 1.15 10 W10s

P ! ! "

2. A block slides down a frictionless inclined plane of height h = 1 m, making angle # with the horizontal. At the bottom of the plane, the block continues to move on a flat surface with a coefficient of friction $ = 0.30. How far does the mass move on the flat surface? We apply the law of Conservation of energy. The equation is E K U% ! % & % The velocity of the block is zero both in the beginning and the end. Thus, 0K% ! We also have friction kE W m g d$% ! ! ' The change in GPE is U mg% ! ' h Putting it all together

1m 3.3m0.30

k

k

E K Um g d mghhd

$

$

% ! % & %' ! '

! ! !

3. A cyclist approaches the bottom of a hill at a speed of 11 m/s. The hill is 6 m high. Ignoring friction, how fast is the cyclist moving at the top of the hill? Assume that he doesnt peddle and ignore air resistance. Since there is no friction, the total mechanical energy is conserved. Thus, . 0U K% & % ! and 0U mgh mgh% ! ' ! 2 21 12 2f iK mv mv% ! ' , Substitute these into the energy conservation equation

2 21 1

2 2

2 2

0

2f i

f i

mgh mv mv

v v gh

& ' !

! '

Thus, 2 22 (11m/s) 2(9.81m/s )(6m)f iv v gh! ' ! '

2 = 2 23.28m /s 1.81m/s! .

4. A 100 kg mass traveling with a velocity of 15 m/s on a horizontal surface strikes a spring with a spring constant k = 5 N/m. a. Find the compression of the spring required to stop the mass if the surface is frictionless. Since there is no dissipative force, we will use the Law of Conservation of Energy. . 0U K! " ! # 2 21 12 20U kx kx! # $ #

2 21 12 20K mv mv! # $ # $

2 21 12 2 0kx mv$ # Solve for x:

mx vk

#

100kg (15m/s) 67.1m5N/m

x # #

b. Find the compression of the spring if the surface is rough ( 0.4k% # 0). The more general form of the equation including nonconservative forces is E U K! # ! " ! kE F d m g x%! # & # $

!!

212U kx! # and 21

2K mv! # $ Putting it all together we get: 2 21 12 2km g x kx mv%$ # $ where 20.40(100kg)(9.81 m/s ) ( 392.4N)km g x x x%$ # $ # $and 2 21 12 2 (5N/m) (2.5N/m)kx x x# #

2

and 2 21 12 2 (100kg)(15m/s) 11250Nmv # # Simplifying (and leaving off the units to make the quadratic equation easier to read) 2392.4 2.5 11250x x$ # $ In standard form this quadratic equation becomes 22.5 392.4 11250 0x x" $ #

$

Solving by the quadratic formula we obtain 24.8m or 182mx x# # The physical solution here is positive. Thus, x = 25 m.

5. An amusement park roller coaster car has a mass of 250 kg. During the ride, it is towed to the top of a 30 m hill, where it is released from rest and allowed to roll. The car plunges down the hill, then up a 10 m hill and through a loop with a radius of 10 m. Assume that the tracks are frictionless. (Use g = 10 m/s2.) a. What is the Potential Energy of the car at the top of the 30 m hill?

! "! "! "Top of Hill

2

250 kg 10 m/s 30 m

75000 J

GPE m g h#

#

# b. What are the Kinetic Energy and the speed of the car at the bottom of the 30 m hill?

Top of Hill Tot Bottom of Hill

2Bottom of Hill

2 2

75000 J

1 2

2 2 (75000 J) 600 m /s 24 m/s250 kg

GPE E KE

KE m v

KEvm

# # #

#

# # # #

c. What are the Kinetic Energy and the speed of the car at the top of the 10 m hill?

! "! "! "210 m10 m 10 m Tot

10 m Tot 10 m

2 210 m

250 kg 10 m/s 10 m 25000 J

75000 J 75000 J 25000 J 50000 J

2 2 (50000 J) 400 m /s 20 m/s250 kg

GPE m g h

GPE KE EKE E GPE

KEvm

# # #

$ # #

# % # % #

# # # #

d. If the hill makes an angle of 60o with the horizontal and the car takes 15 seconds to be towed up the hill, determine the length of the hill, the velocity of the car, the force required to tow the car up the hill, and the power of the motor pulling the car up the hill.

||

2

30m (sin 60 )30m 34.64m

sin 60sin 60

(2500 N)(sin 60 ) 2165 N = 2200 N

34.64m 2.31m/s = 2.3m/s15s

Work Done 75000J 5000J/s 5000 W15s

or= (m g)(sin 60 )( )=(250 kg)(10 m/s )(sin 60 )(2.31

o

o

o

o

L

L

F Wt

Lvt

Pt

P F v v

#

# #

# &

#

# # #'

# # # #'

# ! !

! "m/s)

= (2165 N) 2.31m/s 5001J/s 5.0kW# #