Work Energy Lecture - Iredell-Statesville · Work Done by a Constant Force ... In lifting an object...

Chapter 10 and 11

Work and Energy

Work Done by a Constant Force

The work done by a constant force is defined as

the distance moved multiplied by the component

of the force in the direction of displacement:

Work for Force at an AngleWork for Force at an Angle

If the force is at an angle to the displacement:

Only the horizontal component of the force does any work (horizontal displacement).

Negative and Positive WorkNegative and Positive Work

The work done may be positive, zero, or negative, depending on the angle between the force and the displacement:


In the SI system, the units of work are joules:

As long as this person does

not lift or lower the bag of

groceries, he is doing no work

on it. The force he exerts has

no component in the direction

of motion.

Typical WorkTypical Work

Example: Pulling a Suitcase

A rope inclined upward at 45o pulls a suitcase through the airport. The tension on the rope is 20 N.

How much work does the tension do, if the suitcase is pulled 100 m?

( ) cosW T x θ= ∆

(20 N)(100 m)cos 45 1410 JW = =o

Note that the same work could have been done by a tension of just 14.1 N by pulling in the horizontal direction.

Example: Perpendicular Force and

WorkA car is traveling on a curved

highway. The force due to friction fs

points toward the center of the circular path.

How much work does the frictional force do on the car?

Zero!

General Result: A force that is everywhere perpendicular to the motion does no work.

Work Done by a Constant ForceWork Done by a Constant Force

If there is more than one force acting on an object, we can find the work done by each force, and also the work done by the net force:

Gravitational Work

In lifting an object of weight mg by a height h, the person doing the lifting does an amount of work W = mgh.

If the object is subsequently allowed to fall a distance h, gravity does work W = mgh on the object.

Example: Loading with a Crane

A 3,000 kg truck is to be loaded onto a ship by a crane that exerts an upward force of 31 kN on the truck. This force, which is large enough to overcome the gravitational force and keep the truck moving upward, is applied over a distance of 2.0 m.

(a) Find the work done on the truck by the crane.

(b) Find the work done on the truck by gravity.

(c) Find the net work done on the truck.

app app (31 kN)(2.0 m) 62 kJyW F y= ∆ = =

2

g (3000 kg)( 9.81 m/s )(2.0 m) 58.9 kJyW mg y= ∆ = − = −

net app g (62.0 kJ) ( 58.9 kJ) 3.1 kJW W W= + = + − =

Positive & NegativeGravitational WorkPositive & NegativeGravitational Work

When positive work is done on an object, its speed increases; when negative work is done, its speed decreases.


Solving work problems:

1. Draw a free-body diagram.

2. Choose a coordinate system.

3. Apply Newton’s laws to determine any

unknown forces.

4. Find the work done by a specific force.

5. To find the net work, either

find the net force and then find the work it

does, or

find the work done by each force and add.


Work done by forces that oppose the direction

of motion, such as friction, will be negative.

Centripetal forces do no

work, as they are always

perpendicular to the

direction of motion.

Work Done by a Varying Force

For a force that varies, the work can be

approximated by dividing the distance up into

small pieces, finding the work done during

each, and adding them up. As the pieces

become very narrow, the work done is the area

under the force vs. distance curve.

Kinetic Energy, and the Work-Energy

Principle

Energy was traditionally defined as the ability to

do work. We now know that not all forces are

able to do work; however, we are dealing in

these chapters with mechanical energy, which

does follow this definition.


Principle

If we write the acceleration in terms of the

velocity and the distance, we find that the

work done here is

We define the kinetic energy:


Principle

This means that the work done is equal to the

change in the kinetic energy:

• If the net work is positive, the kinetic energy

increases.

• If the net work is negative, the kinetic energy

decreases.


Principle

Because work and kinetic energy can be

equated, they must have the same units:

kinetic energy is measured in joules.

Question 1

b) 0.707 va) 0.50 v e) 2.00 vd) 1.414 vc) v

Car 1 has twice the mass of Car 2, but they both have the same kinetic energy. If the speed of Car 1 is v, approximately what is the speed of Car 2?

Example: A Dogsled Race

During your winter break, you enter a “dogsled” race across a frozen lake, in which the sleds are pulled by people instead of dogs. To get started, you pull the sled (mass 80 kg) with a force of 180 N at 40° above the horizontal. The sled moves ∆x = 5.0 m, starting from rest. Assume that there is no friction.

(a) Find the work you do.

(b) Find the final speed of your sled.

total you cos

(180 N)(cos40 )(5.0 m) 689 J

xW W F x F xθ= = ∆ = ∆

= ° =

1 1 12 2 2

total 2 2 2f i fW mv mv mv= − =

2 total2f

Wv

m=

total2 2(689 J)4.15 m/s

(80 kg)f

Wv

m= = =

Example: Work and Kinetic Energy

in a Rocket Launch

JmNyFW thrustthrust

96 100.20cos)500)(100.4(cos ×=°×=∆= θ

JmkgymgyFWs

mgravgrav

95 1074.0180cos)500)(80.9)(105.1(coscos 2 ×−=°×=∆=∆= θθ

A 150,000 kg rocket is launched straight up. The rocket engine generates a thrust of 4.0 x 106 N.

What is the rocket’s speed at a height of 500 m? (Ignore air resistance and mass loss due to burned fuel.)

1 2 9

thrust grav20 1.26 10 JK mv W W∆ = − = + = × s

m

m

Kv 21030.1

2×=

∆=

Example: Pushing a Puck

A 500 g ice hockey puck slides across frictionless ice with an initial speed of 2.0 m/s. A compressed air gun is used to exert a continuous force of 1.0 N on the puck to slow it down as it moves 0.50 m. The air gun is aimed at the front edge of the puck, with the compressed air flow 30o below the horizontal. What is the puck’s final speed?

( ) cos (1.0 N)(0.5 m)cos150 0.433 JW F r θ= ∆ = = −o

1 12 2

1 02 2K mv mv W∆ = − =

2 2

1 0

2 2( 0.433 J)(2.0 m/s) 1.51 m/s

(0.5 kg)

Wv v

m

−= + = + =

Example: Work on an ElectronIn a television picture tube, electrons are accelerated

by an electron gun. The force that accelerates theelectron is an electric force due to the electric fieldin the gun. An electron is accelerated from rest by anelectron gun to an energy of 2.5 keV (2,500 eV) over a distanceof 2.5 cm. (1 eV = 1.60 x 10-19 J)

Find the force on the electron, assuming that it is constant and in the direction of the electron’s motion.

19

14

(2,500 ev)(1.6 10 J/eV) 0

(0.025 m)

1.6 10 N

f i

x

K KF

x

−

−

− × −= =

∆

= ×

x f iF x K K∆ = −totalW K= ∆

Potential Energy

An object can have potential energy (PE or U) by

virtue of its surroundings.

Familiar examples of potential energy:

• A wound-up spring

• A stretched elastic band

• An object at some height above the ground

PE is due to an object position

Potential Energy

In raising a mass m to a height

h, the work done by the

external force is

We therefore define the

gravitational potential energy:

(6-5a)

(6-6)

Potential Energy

This potential energy can become kinetic energy

if the object is dropped.

Potential energy is a property of a system as a

whole, not just of the object (because it depends

on external forces).

If , where do we measure y from?

It turns out not to matter, as long as we are

consistent about where we choose y = 0. Only

changes in potential energy can be measured.

Potential Energy Value

• Discovery Ride Packet due Friday

Example: Pike’s Peak or BustAn 82.0 kg mountain climber is in the

final stage of the ascent of Pike’s Peak, which 4,301 m above sea level.

(a) What is the change in gravitational potential energy as the climber gains the last 100.0 m of altitude? Use PE=0 at sea level.

(b) Do the same calculation with PE=0 at the top of the peak.

[ ]2

( )

(82.0 kg)(9.81 m/s ) (4301 m) (4201 m) 80,400 J

a f i f iU mgy mgy mg y y∆ = − = −

= − =

[ ]2

( )

(82.0 kg)(9.81 m/s ) (0) ( 100 m) 80,400 J

b f i f iU mgy mgy mg y y∆ = − = −

= − − =

Example: A Mountain BarA candy bar called the Mountain Bar

has an energy content when metabolized of 212 Cal = 212 kcal. This is equivalent to 8.87 x 105 J.

If an 81.0 kg mountain climber eats a Mountain Bar and magically converts all of its energy content into gravitational potential energy, how much altitude ∆∆∆∆yshould he be able to gain?

f iU mgy mgy mg y∆ = − = ∆

5

2

(8.87 10 J)1,120 m

(81.0 kg)(9.81 m/s )

Uy

mg

∆ ×∆ = = =

Example: A Falling BottleA 0.350 kg bottle falls from rest from a shelf that is 1.75 m

above the floor.

(a) Find the potential energy of the bottle-Earth system when the bottle is on the shelf.

(b) Find the kinetic energy of the bottle-Earth system just before impact with the floor.

total gW W K= = ∆

( )g i fW mg y y mgh= − =

f imgh K K K= ∆ = −

20 (0.350 kg)(9.81 m/s )(1.75 m) 6.01 Jf iK K mgh= + = + =

Elastic Solids & Restoring Forces

An “elastic” material is one that exhibits a restoring force, a force that acts so that it restores a system to an equilibrium position. Examples are springs and rubber bands.

An elastic material stores potential energy when it is deformed and restores it when it returns to equilibrium.

Microscopically, elastic solids depend on the spring-like bonds that bind atoms in a solid.

rubberband

Potential Energy

Potential energy can also be stored in a spring

when it is compressed; the figure below shows

potential energy yielding kinetic energy.

Potential Energy

The force required to

compress or stretch a

spring is:

where k is called the

spring constant, and

needs to be measured for

each spring. This is

Hooke’s Law

Hooke’s LawThe linear proportionality between

force and displacement is found to be valid whether the spring os stretched or compressed, and the force and displacement are always in opposite directions.

Therefore, we write the force-displacement relation as:

( )sp sF k s= − ∆

This relation for the restoring force of a spring is sometimes called Hooke’s Law, named after Robert Hooke, a contemporary of Newton. It is not really a law or nature, but rather a rule of behavior for most springs.

Potential Energy

The force increases as the spring is stretched or

compressed further. We find that the potential

energy of the compressed or stretched spring,

measured from its equilibrium position, can be

written:(6-9)

Work and SpringsWork and Springs

The force needed to stretch a spring an amount x is F = kx.

Therefore, the work done in stretching the spring is

Stretching a Spring

The unloaded spring has a length L0. Hang a weight of mass m on it and it stretches to a new length L. Repeat the process and measure ∆∆∆∆s=L-L0 vs. the applied force Fsp=mg.

We find that Fsp=k∆∆∆∆s, where k is the “spring constant”.

Hooke’s Law and Work

by spring 1 2 1 2

1 12 2

1 22 2

W A A A A

k x k x

= + = −

= −

1 2

2 fW kx=

Question 1

The force vs. displacement curves of three springs are measured.Which spring has the largest spring constant?

a) Spring 1 b) Spring 2 c) Spring 3 d) They are all the same

Question 2

A spring-loaded gun shoots a plastic ball with a speed of 4.0 m/s.

If the spring is compressed twice as far, what is the ball’s speed?

a) 2.0 m/s b) 4.0 m/s c) 8.0 m/s d) 16.0 m/s e) 32.0 m/s

4.0 m/s

Example: Work Done

on a Block by a SpringA 4.0 kg block on a frictionless surface is

attached to a horizontal spring with k = 400 N/m. The spring is initially compressed to 5.0 cm.

(a) Find the work done on the block by the spring as the block moves from x = x

1= -5.0 cm

to its equilibrium position of x = x2= 0 cm.

(b) Find the speed of the block at x2= 0 cm.

1 12 2 2 2

2 2

2f i f i

WW mv mv v v

m= − ⇒ = +

2 2(0.50 J)0.50 m/s

(4.0 kg)f

Wv

m= = =

1 12 2

1 22 2

1 2 2

2(400 N/m)[( 0.05 m) (0 m) ] 0.50 J

xW F x k x x k x k x= ∆ = − ∆ = −

= − − =

∑ ∑

Example: Dragging a Block (1)

A spring is attached to a 2 kg block. The other end is pulled by a motorized toy train that moves forward at 5.0 cm/s. The spring constant is k=50 N/m and the coefficient of static friction between the block and the surface is µµµµs=0.6. The spring is in equilibrium at t=0 s when the train starts to move.At what time does the block

start to slip?

Example: Dragging a Block (2)

net( ) ( ) ( ) 0x sp x s x sp sF F f F f= + = − =∑

s s spf mg F k xµ= = = ∆

2(0.60)(2.0 kg)(9/80 m/s )

(50 N/m)

0.235 m 23.5 cm

smgx

k

µ∆ = =

= =

(23.5 cm)4.7 s

(5.0 cm/s)

xt

v

∆= = =

This is an example of “stick-slip motion”, which is common in nature. Example: behavior of rocks during seismic activity and earthquakes.

Conservation of Mechanical EnergyConservation of Mechanical Energy

Definition of mechanical energy:

(8-6)

Using this definition and considering only conservative forces, we find:

Or equivalently:

Conservation of

Mechanical Energy

mechE K U= +

mech 0E K U∆ = ∆ + ∆ = (Conservation of Mechanical Energy)

Note that ; the correct statement is K U K U∆ ≠ ∆ ∆ = −∆

Conservative and Nonconservative Forces

If friction is present, the work done depends not

only on the starting and ending points, but also

on the path taken. Friction is called a

nonconservative force.

Nonconservative ForcesNonconservative Forces

In the presence of nonconservative forces, the total mechanical energy is not conserved:

Solving,

(8-9)


Potential energy can

only be defined for

conservative forces.


Therefore, we distinguish between the work

done by conservative forces and the work done

by nonconservative forces.

We find that the work done by nonconservative

forces is equal to the total change in kinetic and

potential energies:

Mechanical Energy and Its Conservation

If there are no nonconservative forces, the sum

of the changes in the kinetic energy and in the

potential energy is zero – the kinetic and

potential energy changes are equal but opposite

in sign.

This allows us to define the total mechanical

energy:

And its conservation:

Conservation of Mechanical EnergyConservation of Mechanical EnergyEnergy conservation can make kinematics

problems much easier to solve:

Problem Solving Using Conservation of

Mechanical Energy

In the image on the left, the total

mechanical energy is:

The energy buckets (right)

show how the energy

moves from all potential to

all kinetic.


Mechanical Energy

If there is no friction, the speed of a roller

coaster will depend only on its height

compared to its starting height.


Mechanical Energy

For an elastic force, conservation of energy tells

us:

Other Forms of Energy; Energy

Transformations and the

Conservation of Energy

Some other forms of energy:

Electric energy, nuclear energy, thermal energy,

chemical energy.

Work is done when energy is transferred from

one object to another.

Accounting for all forms of energy, we find that

the total energy neither increases nor

decreases. Energy as a whole is conserved.

Energy Conservation with Dissipative

Processes; Solving Problems

If there is a nonconservative force such as

friction, where do the kinetic and potential

energies go?

They become heat; the actual temperature rise of

the materials involved can be calculated.

Energy Conservation with Dissipative

Processes; Solving Problems

Problem Solving:

1. Draw a picture.

2. Determine the system for which energy will

be conserved.

3. Figure out what you are looking for, and

decide on the initial and final positions.

4. Choose a logical reference frame.

5. Apply conservation of energy.

6. Solve.

Example: Graduation FlingAt the end of a graduation

ceremony, the graduates fling their caps into the air. Suppose a 0.120 kg cap is thrown straight upward with a speed of 7.85 m/s and that frictional forces can be ignored.

(a) Use kinematics to find the speed of the cap when it has risen 1.18 m above the fling point.

(b) Show that the total mechanical energy of the cap is unchanged.

2 2

0 2y y yv v a y= + ∆

2 2 2

0 2 (7.85 m/s) 2( 9.81 m/s )(1.18 m) 6.02 m/sy y yv v a y= + ∆ = + − =

1 12 2

2 20 (0.120 kg)(7.85 m/s) 3.70 Ji i i i iE U K mgy mv= + = + = + =

12 2

2(0.120 kg)(9.81 m/s )(1.18 m) (0.120 kg)(6.20 m/s) 3.70 Jf f fE U K= + = + =

Example: Catching a Home RunExample: Catching a Home Run

At the bottom of the 9th inning,a player hits a 0.15 kg baseball overthe outfield fence. The ball leaves the bat with a speed of 36.0 m/sand a fan in the bleachers catchesit 7.2 m above the point where itwas hit. Neglect air resistance.

(a) What is the kinetic energy Kf of the ball when caught?

(b) What is the speed vf of the ball when caught.

1 12 2

2 20 (0.15 kg)(36 m/s) 97 Ji i i iE U K mgy mv= + = + = + =

2(0.15 kg)(9.81 m/s )(7.2 m) 11 Jf fU mgy= = =

(97 J) (11 J) 86 Jf fK E U= − = − =2 2(86 J)

34 m/s(0.15 kg)

f

f

Kv

m= = =

Speed and PathEnergy is a scalar.

The speed of the cap is vi at height yi and its speed is vf at height yf, independent of the path between the two heights.

Thus the angle at which the cap is launched does not change this result, as long a vi is large enough to carry the cap to height yf.

Question 1

2m

m

h

When a ball of mass m is dropped from height h, its kinetic energy just before landing is K.

If a 2nd ball of mass 2m is dropped from height h/2, what is its kinetic energy just before landing?

(a) K/4 (b) K/2 (c) K (d) 2K (e) 4K

Basic Energy Model

1. There are (at least) two kinds of energy, the kinetic energy K associated with motion of a particle and the potential energy U associated with its position .

2. Kinetic energy can be transformed into potential energy, and potential energy can be transformed into kinetic energy.

3. Under some circumstances, the mechanical energy Emech= K + U is a conserved quantity. Its value at the end of a process is the same as at the beginning. (Energy loss≈0)

Q1: Under what circumstances is Emech conserved?Q2: What happens to the energy when Emech is not conserved?Q3: How do you calculate U for forces other than gravity?

Example: Find the Diver’s DepthExample: Find the Diver’s DepthA 95.0 kg diver steps off a diving

board and drops into the water, 3.00 m below. At some depth d below the water’s surface, the diver comes to rest.

If the nonconservative work done on the diver is Wnc = −5,120 J, what is the depth d?

0iE mgh mgh= + =

( ) 0fE mg d mgd= − + = −

nc f iW E E E mgd mgh= ∆ = − = − −

( ) / 2.49 mncd W mgh mg= − + =

Example: Judging a PuttExample: Judging a PuttA golfer badly misjudges a putt,

giving the ball an initial speed v1, which sends the ball a distance dthat is only one quarter of the distance to the hole.

If the nonconservative force Fdue to the resistance of the grass is constant, what initial speed v2would have been needed to putt the ball from its initial position to the hole?

1 2

1 120f iE K K mv Fd∆ = − = − = −

1 2

2 22(4 )E mv F d∆ = − = −

2 12v v∴ =

1

Example: Landing with a ThudExample: Landing with a ThudA block of mass m1 = 2.40 kg is

on a horizontal table with a coefficient of friction µµµµk = 0.450 between them and is connected to a hanging block of mass m2 = 1.80 kg as shown. When the blocks are released, they move a distance d = 0.50 m, and then m2 hits the floor.

Find the speed of the blocks just before m2 hits.

1 2 1 2; 0;i i iU m gh m gd K E m gh m gd= + = = +

1 1 1 12 2 2 2

1 2 1 2 1 1 22 2 2 2(0); ;f f fU m gh m g K m v m v E m gh m v m v= + = + = + +

1 2( 1 2 22

)f iE E E m m v m gd∆ = − = + −

1 2(1 1 2 2 12

; )nc k k nc kW f d m gd E W m m v m gd m gdµ µ= − = − ∆ = ⇒ + − = −

( ) [ ]2

2 1

1 2

2(9.81 m/s )(0.50 m) (1.80 kg) (0.45)(2.40 kg)21.30 m/s

(2.40 kg) (1.80 kg)

kgd m mv

m m

µ −−= = =

+ +

Example: Marathon ManExample: Marathon ManAn 80.0 kg jogger starts

from rest and runs uphill into a stiff breeze. At the top of the hill the jogger has done work Wnc1 = +18,000 J, air resistance has done workWnc2 = −4420 J, and thejogger’s speed is 3.50 m/s.

Find the height of the hill.1 2

20;i i i f f fE U K E U K mgh mv= + = = + = +

1 2

1 2 2nc nc ncW W W E mgh mv= + = ∆ = +

( )1 2

1 2 2

1 2 2

2

/

(18000 J) ( 4420 J) (80.0 kg)(3.50 m/s) / (80.0 kg)(9.81 m/s )

16.780.0 m

nc nch W W mv mg= + −

= + − − =

Potential Energy Curvesand Equipotentials

Potential Energy Curvesand Equipotentials

The curve of a hill or a roller coaster is itself essentially a plot of the gravitational potential energy:

Potential Energy CurvePotential Energy CurveThe potential energy U and kinetic energy K add to the total

energy E0 (dashed line) at all x values. K vanishes at A and B, which are the turning points of the motion.

Example: A Potential ProblemExample: A Potential ProblemA 1.60 kg object in a

conservative system moves along the x axis, where the potential energy is as shown. A physical example would be a bead sliding along a wire shaped like the red curve.

If the object’s speed at x = 0 is 2.30 m/s, what is its speed at x = 2.00 m?

1 12 2

2 29.35 J; (1.60 kg)(2.30 m/s) 4.23 J; 13.58 Ji i i i fU K mv E E= = = = = =

1 2

24.15 J;f f f f fU K mv E U= = = −

[ ]2( ) 2 (13.58 J) (4.15 J)3.43 m/s

(1.60 kg)

f f

f

E Uv

m

− −= = =

Power

Power is the rate at which work is done –

The difference between walking

and running up these stairs is

power – the change in

gravitational potential energy is

the same.

(6-17)

In the SI system, the units of

power are watts:

PowerPowerPower is a measure of the rate at which work

is done:

SI power unit: 1 J/s = 1 watt = 1 W

1 horsepower = 1 hp = 746 W

Power and VelocityPower and VelocityIf an object is moving at a constant speed in

the presence of friction, gravity, air resistance, and so forth, the power exerted by the driving force can be written:

(7-13)

Power

Power is also needed for acceleration and for

moving against the force of gravity.

The average power can be written in terms of the

force and the average velocity:

(6-17)

PowerPower

Example:

The Power of a MotorA small motor is used to operate a lift that raises a load of

bricks weighing 500 N to a height of 10 m in 20 s at constant speed. The lift weighs 300 N.

What is the power output of the motor?

( )cos cos 0P F v Fv Fv Fvφ= ⋅ = = =r r

(500 N 300 N)(10 m/20 s) 400 W 0.54 hpP = + = =

Energy

• Energy (W) lost or gained by any system

is determined by:

W = Pt

• Since power is measured in watts (or

joules per second) and time in seconds,

the unit of energy is the wattsecond (Ws)

or joule (J)

Energy

• The watt-second is too small a quantity for most practical purposes, so the watt-hour(Wh) and kilowatt-hour (kWh) are defined as follows:

• The killowatt-hour meter is an instrument used for measuring the energy supplied to a residential or commercial user of electricity.

1000

(h) time (W) power(kWh)Energy

×=

(h) time (W) power (Wh)Energy ×=

Applications

• Household wiring

• Most older homes, without electric heating,

have a 100 A service.

• Power is broken down into different circuits

utilizing 15 A, 20 A, 30 A and 40 A protective

breakers.

• Maximum load on each breaker should not

exceed 80% of its rating (12 A of a 15 A

circuit breaker).

Typical wattage ratings of some common

household items

Insert Table 4.1Insert Table 4.1

Mechanical Advantage

If you hold the wrench here, you need a lot of force...

If you hold the wrench here, you don’t need as much force...

…but you don’t move very far.

…but your hand moves a long way.


It takes the same amount of work to turn the bolt.

You can opt for a lot of force and little distance.W = F x d

Or you can choose a little force but a lot of distance.W = f x D

In many of our machines, we want to increase our force, so we don’t mind going the extra distance.


Question: Which ramp would you prefer to use to move a heavy weight to the top of the box?

Answer: This ramp requires less force, but you have to move the weight a longer distance.


With a gear box, you were able to create a large torque here...

…but you had to turn this handle many times.Remember: W = τ x θ

Machines

• A machine eases the load by changing

either the magnitude to the direction of a

force to match the force to the capability of

the machine or person.

What is a Simple Machine?

• A simple machine has

few or no moving

parts.

• Simple machines

make work easier

Wheels and Axles

• The wheel and axle are a simple machine

• The axle is a rod that goes through the wheel which allows the wheel to turn

• Gears are a form of wheels and axles

Pulleys

• Pulley are wheels and

axles with a groove

around the outside

• A pulley needs a

rope, chain or belt

around the groove to

make it do work

Inclined Planes

• An inclined plane is a

flat surface that is

higher on one end

• Inclined planes make

the work of moving

things easier

Wedges

• Two inclined planes joined back to back.

• Wedges are used to split things.

Screws

• A screw is an inclined plane wrapped around a shaft or cylinder.

• The inclined plane allows the screw to move itself when rotated.

Levers-First Class

• In a first class lever

the fulcrum is in the

middle and the load

and effort is on either

side

• Think of a see-saw

Levers-Second Class

• In a second class

lever the fulcrum is at

the end, with the load

in the middle

• Think of a

wheelbarrow

Levers-Third Class

• In a third class lever

the fulcrum is again at

the end, but the effort

is in the middle

• Think of a pair of

tweezers

Simple Machines

• Simple Machines can

be put together in

different ways to

make complex

machinery

• Machines make work easier by changing

the size and/or the direction of the force

– In order for work to be done:

• A force must be exerted

• There must be movement in direction of the force

• The two most common forces being

overcome are friction and gravity

• There are 2 general forces involved when

using a machine to do work

– Effort (Input) Force (FE) – This is the force that

is applied to the machine

– Resistance (Output) Force (FR) – This is the

force that the machine applies

• There are also two types of work involved

– Work In (Win) – The work done on the

machine (energy put into the machine)

– Work Out (Wout) – The work done by the

machine (energy put out by the machine)

Since work is force x distance, than we can find the value

of the work in and the work out

Win=Fe x de Wout=Fr x dr

In an “ideal machine”, (no friction) no energy would be lost

so work in and work out would be equal

Win=Wout Fe x de = Fr x dr

In reality, work out will always be less than work in because

energy is lost through friction

Mechanical Advantage• Most machines multiply or increase the force applied to it

so that the resistance force is greater than the effort

force (Fr>Fe)

• In order for the work in to still equal the work out (Win =

Wout), the resistance force must travel a shorter distance

than the effort force (dr<de)

• The number of times that a machine multiplies the effort

force is called the Mechanical Advantage (MA).

MA=Fr/Fe

• the Ideal Mechanical Advantage (IMA). IMA=de/dr

Machine Efficiency• Most machines not all input work goes to output work

• The efficiency of a machine is the percentage ratio of

output to input work (value is less than 100%)

• e = (Wout / Win ) * 100

• It can also be defined as the percentage ratio of the

meacnical advantage to the ideal mechanical advantage

• e = (MA / IMA ) * 100

Compound Machines

• A compound machine is a

combination of simple

machines that work together.

Human Walking Machine

• A human is a combination of

simple machines that work

together.

• Rigid Bar (bone)

• Source of Force (muscle)

• Pivot (joint)

• Resistance (body or weight

lifted)

Summary of Chapter 10 and 11

•Work:

•Kinetic energy is energy of motion:

• Potential energy is energy associated with forces

that depend on the position or configuration of

objects.

•

•The net work done on an object equals the change

in its kinetic energy.

• If only conservative forces are acting, mechanical

energy is conserved.

• Power is the rate at which work is done.

Work Energy Lecture - Iredell-Statesville · Work Done by a Constant Force ... In lifting an object...

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Transcript of Work Energy Lecture - Iredell-Statesville · Work Done by a Constant Force ... In lifting an object...