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Applied Natural Sciences

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Copyright © 2012 Pearson Education Inc.

PowerPoint® Lectures forUniversity Physics, Thirteenth Edition

– Hugh D. Young and Roger A. Freedman

Lectures by Wayne Anderson

Chapter 6

Work and Kinetic Energy

The “money” of physics

LEARNING GOALS

• What it means for a force to do work on a body, and how

to calculate the amount of work done.

• The definition of the kinetic energy (energy of motion) of

a body, and what it means physically.

• How the total work done on a body changes the body’s

kinetic energy, and how to use this principle to solve

problems in mechanics.

• How to use the relationship between total work and

change in kinetic energy when the forces are not

constant, the body follows a curved path, or both.

• How to solve problems involving power (the rate of doing

work).4

Primitive Economics

Do your job Get paid

5

Modern Economics

Using money simplifies economics and accounting.

Do your jobGet paid

Buy stuff

6

Why Energy Helps

Motion, in general, is hard to calculate.

Using forces, momentum, acceleration, etc. gets complicated because they are all vectors (have 

magnitude & direction).

Energy is not a vector; it’s just a number.

Can predict motion by figuring out how much energy that motion will “cost.”

7

PART 2: Example of skier

What is the speed of the skier downhill?(no friction)

h Answer:

Follows 'fast' from consideration of energy

Newton:

8

Where are we

Classical mechanics

is ready!!Kinematics

Newton’s laws

Why continue?

9

Where are we

Problem can be solved often more convenient

Use laws derived from Newton’s laws

Laws around (conservation) energy

Laws around (conservation) momentum

Classical mechanics

is ready!!Kinematics

Newton’s laws

10

Relation Newton’s laws and Energy

Newton’s laws

Energy

Dutch: arbeidWork

11

The definition of work, when the force is parallel to the displacement:

SI work unit:newton-meter (N·m) = joule, J

Work done by constant force

12

13

If the force is at an angle to the displacement:

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

Work done by force at angle

14

Work Summary

W F x

cosxW F x F x

SI Units for work:

1 joule = 1 J = 1 N·m

1 electron-volt = 1 eV = 1.602 x 10-19 J

16

The work can also be written as the dot product of the force F and the displacement d:

Work done by constant force

17

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

Negative and Positive Work

18

Unbanked curves

What is the maximum speed for friction coefficient μ?

1) Negative2) 03) Positive

How much is work is done by friction?

19

A 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.

Perpendicular Force and Work

20

Work (more general)

22

Work (more general)

22 2 2

1 1 1 1

yP x z

x y zP x y zW F dl Fdx F dy Fdz

23

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

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

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

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Gravitational WorkIn 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.

W mgh

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

Positive & Negative Gravitational Work

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General definition

Work done by a force on an object :

F

dr

W Fdr

r1

r2

• result: scalar (unit J=Nm)

• dimension of energy

Also useful as:

• force is unknown

• angle is not constant

• motion is not along straight line

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Work

A woman holding a bowling ball still in her hand. The work she performed on the ball

2. can not be calculated with this information.

1. depends on the mass of the ball.

3. is zero.

Answer: 3. There is force needed, but there is no movement, and so there is no work done.

28

Work by gravitational force

v

A comet is approaching Earth. For the labor WA the earth during the approach to the comet is doing and the work WK done by the comet on the earth:

2. WA > 0 and WK > 0

1. WA > 0 but WK < 0

3. WA < 0 but WK > 0

4. WA < 0 and WK < 0

Answer: 2. For both forces are the force vector and motion vector in the same direction

29

FKA

FAK

sK

sA F

s 0

F

s 0

At equal displacement(contact forces):

Two objects: WAB = - W BA

Gravitational force is

not a contact force:

WAB ≠ - W BA

30

Magnitude of the force not constant: spring

Force on spring: F = k x Hooke’s law

elastic deformations

By hand on spring:

By spring on hand

Labour can be positive or negative

Two objects in contact: WAB = - W BA

0

Work done by spring:

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Law work‐energy

Due to force speed boot increases

Constant force constant acceleration2 2

2 12v v as

2 2

2 1( )2

v vW Fs mas m ss

2 2

2 1

1 12 2

W mv mv

2 1W K K K

Newton: F = ma, and work

hence:

Definition: K = ½ m v2, kinetic energy

Work-energy

32

Work –energy more general

2 2

2 1

1 12 2

W mv mv

2 2

1 1( )

x x

x x

dvW F x dx m dxdt

dv dv dx dvvdt dx dt dx

2 2

1 1

x v

x v

dvW mv dx mvdvdx

2 1W K K K

Work = difference in kinetic energy33

Kinetic energy for various objects

34

Law of Work and Kinetic Energy

The work done by the net force equals the change in kinetic energy

W K K2 K1

thus follows directly from Newton's laws

Apply:

speed / velocity change given position-dependent force

force / displacement given speed / speed change

35

Problem Solving Strategy

Picture: The way you choose the +y direction or the +x direction can help you to easily solve a problem that involves work and kinetic energy.

Solve:1. Draw the particle first at its initial position and second at its final position. For convenience, the object can be represented as a dot or box. Label the initial and final positions of the object.2. Put one or more coordinate axes on the drawing.3. Draw arrows for the initial and final velocities, and label them appropriately.4. On the initial-position drawing of the particle, place a labeled vector for each force acting on it.5. Calculate the total work done on the particle by the forces and equate this total to the change in the particle’s kinetic energy.

Check: Make sure you pay attention to negative signs during your calculations. For example, values for work done can be positive or negative, depending on the direction of the displacement relative to the direction of the force. Kinetic energy values, however, are always positive.

36

Example: A Dogsled RaceDuring your winter break, you enter a

“dogsled” race across a frozen lake, in which the sleds are pulled by students 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)(cos 40 )(5.0 m) 689 JxW W F x F x

1 1 12 2 2total 2 2 2f i fW mv mv mv

2 total2f

Wvm

total2 2(689 J) 4.15 m/s(80 kg)f

Wvm

37

Example:  Work and Kinetic Energy in a Rocket Launch

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.)

6 9thrust thrust ( ) (4.0 10 N)(500 m) 2.0 10 JW F y

4 2 9grav ( ) ( ) (1.5 10 kg)(9.80 m/s )(500 m) 0.74 10 JW w y mg y

1 2 9thrust grav2 0 1.26 10 JK mv W W 2 129.6 m/sKv

m

38

Typical example of excercise

Have to calculate friction force

Typical Energy/Work:

• we know begin/end speed

• distance travelled

2 object: can be treated seperately

The blocks in the first figure initially move at a speed v = 0.9 m/s to the right / down, but after a distance s = 2.0 m come to a halt. Calculate the coefficient of friction between block and table.

41

setup

FtFw

m1g

N Ft

m2g

y

x

d ( )w tF F s F r

2d ( )tm g F s F r

21120K m v 21

220K m v

1w k kF N m g

m1

m2

v

v

42

y

x

2112 ( )w tm v F F s

212 22 ( )tm v m g F s

d ( )w tF F s F r

2d ( )tm g F s F r

21120K m v 21

220K m v

211 2 2 2 12 ( ) ( ) ( )w km m v m g F s m g m g s

+

211 2 22

1

( ) 0.79km m v m gs

m gs

dW K F r

Ft elimineren

1w kF m g

execute

43

Power 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

James Watt (1736-1819)

Power

44

Power and Velocity

so v v tt

W F F v t

SI Units for power: 1 watt = 1 W = 1 J/s

341 hp = 550 ft lb/s = 746 W kW

61 kW h = (1000 W)(3600 s) = 3.6 10 W h = 3.6 MJ

WP F vt

Power is the rate of energy flow.

45

46

Example:The Power of a Motor

A 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 Fv

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

47

Summary

48

Summary

49

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