Chap. 6: Work and

Kinetic Energy

1. What is Work? Work as Energy Transfer Work done by a (constant/varying) force (scalar product & integrals)

2. Kinetic Energy A way to describe the status of the motion Work-Energy Theorem

1

Case Study: Constant Force

Work and Energy

Math

• Integrals

• Scalar product

of two vectors

2

Work and Energy

Math

• Integrals

• Scalar product

of two vectors

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Work and Energy

What is “Work”?

6

Work and Energy

1) Effort or labor to get something done. Solving problems is hard work.

2) A person’s job. What kind of work do you do?

3) A piece of music, a painting, or sculpture. A work of art.

4) A scalar quantity involving both a force and a distance. When you lift a barbell against Earth’s gravity, work is done. The

heavier the barbell or the higher we lift the barbell, the more work is done on the barbell.

What is “Work”?

(force) (distance)

(force)

(distance)

W

W

W

7

PHYS218

Work and Energy

4) A scalar quantity involving both a force and a distance. A weight lifter who holds a barbell weighting 1000 N does no

work on the barbell. (∵)Zero distance.

Won barbel = 0 But he may get really tired doing so.

Work may be done on the muscles by stretching and contracting,

which is force times distance on a biological (small) scale.

Won muscles 0

8

Work and Energy

Work = Energy Transfer

[Thermodynamics]

Status 1

Status 2

Status 1

Status 2

9

Work and Energy

Work (W) Done by

a Constant Force (I)

A tractor is doing work on (or “energy transfer” to) a sled

of firewood as it exerts the force over a displacement.

11

Engine

Energy Transfer

Work and Energy

[Hint] Both magnitude (F) and

directions (q ) must be taken

into account.

FN

Ff = 3500 N

FG = 14,700 N

F.B.D. on sled

d = 20 m

14

Work (W) Done by

a Constant Force (I)

[Q] Is the force effectively pulling a sled of firewood?

(distance)

cos

(force)energy kinetic of difference

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Work and Energy

Ff = 3500 N

FG = 14,700 N

FN

m

Fa

d = 20 m

18

Work and Energy

(distance)

cos

(force)energy kinetic of difference

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[Q] What is this?

m

Fa

23

[A] That is a sum of Wtension and Wfriction (= Wnet).

Work-Energy Theorem

Work and Energy

Ff = 3500 N

FG = 14,700 N

FN

[Q] Find vf if vi = 2.0 m/s using

“work-energy theorem”.

Problem-solving steps:

(1) F.B.D. for an object

(2) Wi = Fi d cosqi = F||d (if all forces are constant)

3 Wnet SWi

4 Wnet DK

Analyze the motion

d = 20 m

24

NOTE: You have noticed WFN and WFG are zero.

“No Work”

What was a special condition to see this?

Work and Energy

Special Case: “No Work”

[Example] “Work” done on the

bag by the person..

Special case: W = 0 J

a) WP = FP d cos ( 90o )

b) Wg = m g d cos ( 90o )

Nothing to do with the motion

25

Work and Energy

Work Energy Theorem

Wnet = Kf – Ki

You measure Force(s) and Distance. You measure Velocities.

26

Work and Energy

e.g., Work by a Baseball Pitcher

A baseball pitcher is doing work on (or

“energy transfer” to) the ball as he

exerts the force over a displacement.

vi = 0

Ki = (1/2) m vi 2

vf = 44 m/s

Kf = (1/2) m vf 2

Initial Status Final Status 27

Work and Energy

A 50.0-kg crate is pulled 40.0 m by a constant force

exerted (FP = 100 N and q = 37.0o) by a person. Assume

a coefficient of friction force mk = 0.110. Determine the

work done by each force acting on the crate and its net

work. Find the final velocity of the crate if d = 40 m and

vi = 0 m/s.

Example 1

28

Work and Energy

Example 1 Solution

180o

90o

d

F.B.D.

29

Work and Energy

Example 1 Solution (cont’d)

Wnet = SWi

= 1302 [J] (> 0)

The body’s speed increases.

32

Work and Energy

Example 2

(4) W-E Theorem to

find v2.

(1) F.B.D.

(2) W by each force

(3) Wnet

33

Work and Energy

Example 2 Solution

(1) F.B.D.

(2) W by each force

FN

FG

Ff

FP

34

Work and Energy

WP = FP d /cos(30o)

WN = FN d

FN = FG cos(30o) or FG

Wf = Ff d cos(30o)

WG = FG d cos(90o)

Example 2 Wrong Solution

(1) F.B.D.

(2) W by each force

FN

FG

FP or FP/cos(30o)

Ff

40

Work and Energy

Example 2 Wrong Solution

(4) W-E Theorem to find v2 :

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41

Chap. 6: Work and

Kinetic Energy

1. What is Work? Work as Energy Transfer Work done by a (constant/varying) force (scalar product & integrals)

2. Kinetic Energy A way to describe the status of the motion Work-Energy Theorem

42

Case Study: Varying Force

Work and Energy

W DK = Kb - Ka = ½ m vb2 ½ m va

2

“Work-Energy Theorem”

Work

Distance, d

Work: Area in F-d Graph

[F = constant, q = constant]

Work and Energy

Work Done by Work Done by

a Constant Force (I)a Constant Force (I)

Work (W)

Both magnitude (F) and

directions (q ) must be

taken into account.

W [Joule] = ( F cos q ) d = F d

How effective is the force in moving a body?

a b F

cosq

An ideal scenario …

43

Work and Energy

W DK = Kb - Ka = ½ m vb2 ½ m va

2

“Work-Energy Theorem”

Work

Distance, d Work and Energy

Work Done by Work Done by

a Constant Force (I)a Constant Force (I)

Work (W)

Both magnitude (F) and

directions (q ) must be

taken into account.

W [Joule] = ( F cos q ) d = F d

How effective is the force in moving a body?

Fco

sq

Work: Area in F-d Graph [F = varied, q = constant]

She is exhausted.

If we know how to calculate the area in F-d graph, we are okay! 44

Work and Energy

Fco

sq

lb

W = F|| dl

la

Motion ab is

collection of straight-

line motion with a

constant force

46

Work Done by a Varying Force (direction and magnitude)

Work and Energy

Spring Force (Hooke’s Law) [q = constant, F = varied]

FS(x) = k x

FP FS

Natural Length x > 0

x < 0

Spring Force

(Restoring Force):

The spring exerts

its force in the

direction opposite

the displacement.

x 0

47

Work and Energy

Work Done on Box by FS

FS(x) = – k x

Natural Length

FP FS

xf

WS = FS(x) d x = – ½ k xf2 + ½ k xi

2

xi

Final position

Initial position

48

Work and Energy

x2

WP = FP(x) dx

x1

Work Done “to Stretch a Spring”

FP(x) = k x

W

Natural Length

FP FS

xf

WP = FP(x) d x = ½ k xf2 – ½ k xi

2

xi

Initial position

Final position

49

Work and Energy

Problem Solving Steps [Varying Force]

ifnet

net

4

3

d 2)

object eachfor F.B.D. 1)

2

1

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i

i

l

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ii

50

Work and Energy

Example 3

A person pulls on the spring, stretching it

3.0 cm, which requires a maximum force

of 75 N. How much work does the person

do ? If, instead, the

person compresses

the spring 3.0 cm,

how much work

does the person do ?

51

Example 3 Solution

52

Work and Energy

Example 4

A person pulls on the spring, stretching it

3.0 cm, which requires a maximum force

of 75 N. How much work does the spring

do ? If, instead, the

person compresses

the spring 3.0 cm,

how much work

does the spring do ?

54

Example 4 Solution

55

Work and Energy

You are weighing 600 N on a bathroom scale containing a stiff

spring. In equilibrium the spring is 1.0 cm under your weight.

Find the spring constant and the work done by the spring.

Example 5

57

Example 5 Solution

58

Work and Energy

Example 6

A 1.50-kg block is pushed against a spring

(k = 250 N/m), compressing it 0.200 m, and

released. What will be the speed of the

block when it separates from the spring at

x = 0? Assume mk =

0.300.

(i) F.B.D. first !

(ii) x < 0

FS = k x

60

Example 6 Solution

61

Work and Energy

Example 7

A 1.50-kg block is pushed against a spring,

compressing it 0.200 m, and released. The

spring force is given as FS(x) = kx + bx2 where

k = 250 N/m and b = 300 N/m2. What will be

the speed of the block when it separates from

the spring at x = 0?

Assume mk = 0.300.

(i) F.B.D. first !

(ii) x < 0

FS = k x+bx2

63

Example 7 Solution

64

Work and Energy

Example 8 Today’s cars have “8 km/h bumpers” that are designed to

elastically compressed and rebound without any physical

damage at speeds below 8 km/h. If the material of the bumper

permanently deforms after a compression of 1.5 cm, but remains

like a elastic spring up to that point, what must the effective

spring constant of the bumper material be, assuming the car has

a mass of 1150 kg and is tested by ramming into a solid wall?

66

1) Set up a model (simplified

version of a physical system) to

explain the phenomena.

2) …

Work and Energy

Today’s cars have “8 km/h bumpers” that are designed to

elastically compressed and rebound without any physical

damage at speeds below 8 km/h. If the material of the bumper

permanently deforms after a compression of 1.5 cm, but remains

like a elastic spring up to that point, what must the effective

spring constant of the bumper material be, assuming the car has

a mass of 1150 kg and is tested by ramming into a solid wall?

Bumper Car

1) Set up a model (simplified

version of a physical system) to

explain the phenomena.

2) Identify …

67

Example 8

Work and Energy

Problem 4: (25 points)

A particle of mass M moves in one dimension under the influence of a force that is given

by:

F(x) = A x + B x2 –C x3

where the direction of motion is taken to be +x and A, B, and C are given constants.

a. (15 pts) The particle starts at position –x0 and moves to position +x0. Express the

work done on this particle by the force given above in terms of A, B, C, and x0.

b. (10 pts) Assume that when the particle is at –x0 it has zero velocity. Find the velocity

of the particle (in terms of A, B, C, x0, and M) when its reaches +x0.

70

Example 9

Work and Energy

Problem 4: (25 points)

A block is sliding on a smooth, flat, horizontal plane with speed v0. Initially, there is no

friction. At some position, which you should call x = 0, the plane starts to get rough, and

friction increases linearly with distance, increasing by 1 in a distance L. This means that the

coefficient of kinetic friction is a function of position, and we can write:

m(x) = x / L .

Thus we have m(0) = 0 and m(L) = 1. Calculate how far into the rough surface the block

slides before coming to rest. Note that L is the distance over which m changes by 1, not the

distance the block travels. The distance the block travels depends on v0, g, and L.

71

Example 10