Chapter 5Chapter 5

Work, Power and EnergyWork, Power and Energy

ObjectivesObjectives

Section 1 – WorkSection 1 – Work

• Recognize the difference between the scientific Recognize the difference between the scientific and ordinary definitions of workand ordinary definitions of work

• Define work by relating it to force and Define work by relating it to force and displacementdisplacement

• Identify where work is being performed in a Identify where work is being performed in a variety of situationsvariety of situations

• Calculate the net work done when many forces Calculate the net work done when many forces are applied to an objectare applied to an object

ObjectivesObjectives Section 2 – EnergySection 2 – Energy

• Identify several forms of energyIdentify several forms of energy

• Calculate kinetic energy for an objectCalculate kinetic energy for an object

• Apply in the work-kinetic energy theorem to solve Apply in the work-kinetic energy theorem to solve problemsproblems

• Distinguish between kinetic and potential energyDistinguish between kinetic and potential energy

• Classify different types of potential energyClassify different types of potential energy

• Calculate the potential energy associated with an Calculate the potential energy associated with an object’s positionobject’s position

ObjectivesObjectives

Section 3 – Conservation of EnergySection 3 – Conservation of Energy

• Identify situations in which conservation Identify situations in which conservation of mechanical energy is validof mechanical energy is valid

• Recognize the forms that conserved Recognize the forms that conserved energy can takeenergy can take

• Solve problems using conservation of Solve problems using conservation of mechanical energymechanical energy

ObjectivesObjectives

Section 4 – PowerSection 4 – Power

• Relate the concepts of energy, time and Relate the concepts of energy, time and powerpower

• Calculate power in two different waysCalculate power in two different ways

• Explain the effect of machines on work Explain the effect of machines on work and powerand power

WorkWork

Work – the product of force and Work – the product of force and displacement, as long as they are displacement, as long as they are both in the same direction.both in the same direction.

W F d

This is a vector times a vector.

In this case, a vector times a vector gives a scalar.

Work is measured in Joules (J) which is the same as a Newton x meter.

ExampleExample

How much work is done on an object if How much work is done on an object if a force of 30 Newtons displaces the a force of 30 Newtons displaces the object 200 meters?object 200 meters?

W F d

(30 )(200 )W N m

6000W J

What happens if the force and displacement are not in the same direction?

F

d

( cos )W F d or

cosW Fd

Where θ is the angle between the two vectors

Note: We’ll talk about where the cosine magically comes from in class.

Example:

As Alex pulls his red wagon down the sidewalk, the handle of the wagon makes an angle of 60o with the pavement. If Alex exerts a force of 100 Newtons along the direction of the handle, how much work is done when the displacement of the wagon is 20 meters along the ground?

100

N

20 m

60o

100

N20 m

60o

( cos )W F d(100 )(cos60 )(20 )oW N m

1000W J

Example:

A constant force of 50 Newtons is applied over a distance of 10 meters. Draw a graph to represent this information. Calculate the work done by this force. Calculate the area under the graph. How do they compare?

d (m)

F (N)

50

10

W F d (50 )(10 )W N m500W J

A b h (10 )(50 )A m N500A J

If given a graph of

Force vs. Displacement

The work can be found by calculating the area under the graph.

That’s enough for now…

Lets assume you own a moving Lets assume you own a moving company that employs two workers.company that employs two workers.

Employee one moves ten boxes and Employee one moves ten boxes and gets the job done by lunch. gets the job done by lunch.

Employee two moves ten boxes but Employee two moves ten boxes but takes his time and finishes by 5pm.takes his time and finishes by 5pm.

Who does more work?Who does more work?

Who would you pay more?Who would you pay more?

While there is no correct answer here, there is an added piece that work

alone does not address.

Time is the added piece.

PowerPower

Power is the rate at which work is Power is the rate at which work is done.done.

WP

t

secJoule

WattUnit:

Example:Example:

If 3000 Joules of work is performed on If 3000 Joules of work is performed on an object in 1.0 minute, what is the an object in 1.0 minute, what is the power expended on the object?power expended on the object?

WP

t

300060J

Ps

50P W

Example:Example:A 200 Newton force is applied to an A 200 Newton force is applied to an

object that moves in the direction of object that moves in the direction of the force. If the object travels with a the force. If the object travels with a constant velocity of 10 m/s, calculate constant velocity of 10 m/s, calculate the power expended on the object.the power expended on the object.

WP

t F d

tF v

(200 )(10 )msP N

2000P W

EnergyEnergy

When work is done on an object, the When work is done on an object, the energy of the object is changed.energy of the object is changed.

Kinetic EnergyKinetic Energy

F ma W FdW ma d

2 22 1 2v v a d

2 22 1

2v v

ad

2 22 1

2v v

W m 2 21 1

2 22 1W mv mv

The work done on the object changes a The work done on the object changes a quantity called kinetic energy that is quantity called kinetic energy that is related to the mass and the square related to the mass and the square of the speed of the object.of the speed of the object.

2 21 12 22 1W mv mv

212KE mv

2 1W KE KE KE

This is known as the work-energy theorem.

Example:Example:

A 10kg object subjected to a 20 Newton A 10kg object subjected to a 20 Newton force moves across a horizontal force moves across a horizontal frictionless surface in the direction of the frictionless surface in the direction of the force. Before the force was applied, the force. Before the force was applied, the speed of the object was 2.0 meters per speed of the object was 2.0 meters per second. When the force is removed, the second. When the force is removed, the object is traveling at 6.0 meters per object is traveling at 6.0 meters per second. Calculate the following quantities: second. Calculate the following quantities: (a) KE(a) KE11 (b) KE (b) KE22 (c) (c) ΔΔKE (d) W and (e) KE (d) W and (e) distance.distance.

2121 1KE mv

2121 (10 )(2.0 )msKE kg

1 20KE J

2122 2KE mv

2122 (10 )(6.0 )msKE kg

2 180KE J

2 1KE KE KE 180 20KE J J 160KE J

W KE160W J

W FdW

dF

16020J

dN

8.0d m

That’s enough for today…

Gravitational Potential EnergyGravitational Potential Energy

Lifting a book from the floor to the Lifting a book from the floor to the table applies a force over a distance table applies a force over a distance therefore work is done on the book therefore work is done on the book but the kinetic energy is unchanged.but the kinetic energy is unchanged.

In this case the force of gravity is In this case the force of gravity is overcome by work.overcome by work.

gF mg W F d W mg d PE mg h

Example:Example:

A 2.0 kilogram mass is lifted to a A 2.0 kilogram mass is lifted to a height of 10 meters above the height of 10 meters above the surface of the Earth. Calculate the surface of the Earth. Calculate the change in the gravitational potential change in the gravitational potential energy of the object.energy of the object.

PE mg h

2(2.0 )(9.8 )(10 )ms

PE kg m

196PE J

Conservation of EnergyConservation of Energy

Energy can not be created or destroyed.Energy can not be created or destroyed.

In the system, the sum of the potential In the system, the sum of the potential energy and kinetic energy will be energy and kinetic energy will be conserved; a change in one is conserved; a change in one is accompanied by an opposite change in accompanied by an opposite change in the other.the other.

Example:Example:

A 2.0 kg ball is tossed with a velocity A 2.0 kg ball is tossed with a velocity of 4.0 m/s straight into the air. How of 4.0 m/s straight into the air. How high does the ball travel upwards? high does the ball travel upwards? Calculate the initial kinetic energy of Calculate the initial kinetic energy of the ball. Calculate the potential the ball. Calculate the potential energy of the ball at it’s peak.energy of the ball at it’s peak.

Givens:1 4.0 msv

29.8 ms

a 0 mstopv

2.0m kg

2 22 1 2v v a d 21 2v a d

21

2v

da

2

2

2

16

2( 9.8 )

ms

ms

d

0.815d m

2121 1KE mv

2121 (2.0 )(4.0 )msKE kg

1 16KE J

2PE mgh

22 (2.0 )(9.8 )(0.815 )ms

PE kg m

2 16PE J

2 0KE J 1 0PE J

PE KE

1 1 2 2PE KE PE KE 1 2E E

or

Roller CoasterRoller Coasterv1 = 4.0m/sm=1.0 kg

h1 = 10 m

h1 = 5 mh1 = 4 m h1 = 2 m

Calculate the total energy and the velocity at the top of each hill.

h1 = 10 m

h1 = 5 mh1 = 4 m h1 = 2 m

v1 = 4.0m/s

m=1.0 kg

TE KE PE 21

2TE mv mgh

221

2 (1 )(4 ) (1 )(9.8 )(10 )m msT s

E kg kg m

106TE J

TE KE PE 21

2TE mv mgh 21

2 Tmv E mgh 2 2Emv gh

22(106 )1 2(9.8 )(5.0 )J mkg s

v m 10.7 msv

TE KE PE 21

2TE mv mgh 21

2 Tmv E mgh 2 2Emv gh

22(106 )1 2(9.8 )(4.0 )J mkg s

v m

11.6 msv

TE KE PE 21

2TE mv mgh 21

2 Tmv E mgh 2 2Emv gh

22(106 )1 2(9.8 )(2.0 )J mkg s

v m

13.2 msv

PendulumPendulum

Simple Harmonic Motion – In the Simple Harmonic Motion – In the absence of friction the pendulum will absence of friction the pendulum will swing back and forth continuously.swing back and forth continuously.

Example:Example:

A pendulum whose bob weighs 12 A pendulum whose bob weighs 12 Newtons is lifted a vertical height of Newtons is lifted a vertical height of 0.40 meter from its equilibrium 0.40 meter from its equilibrium position. Calculate the change in position. Calculate the change in potential energy between maximum potential energy between maximum height and equilibrium height, the height and equilibrium height, the gain in kinetic energy, and the gain in kinetic energy, and the velocity at the equilibrium point.velocity at the equilibrium point.

PE mg h (12 )( 0.40 )PE N m

4.8PE J

KE PE ( 4.8 )KE J 4.8KE J 21

2KE mv2KEmv

2(4.8 )1.2

Jkgv

2.8msv