Work and energy

1. Work done by a constant force

Definition:

1a. Positive and negative work

[W] = N*m = J

Work done by forces that oppose the direction of motion will be negative.

dFFddFW

cos||

Centripetal forces do no work, as they are always perpendicular to the direction of motion.

Units:

1

F

xF

yF

x

y

f

m5

a

Example: An object of unknown mass is displaced 5 m by a constant

force F = 20 N as shown below (angle θ=60º). Force of friction is f = 6 N.

Find the work done by each of these forces and the total work.

JmNNxfFW

JmNfxxfW

JmNxFxFW

tot

fr

xF

20)5(660cos20cos

30 )5)(6( )180(cos

50)5)(60)(cos20()cos(

0

0

0

Find mass of the object if the coefficient of kinetic friction is 0.5. 2

A. Positive

B. Negative

C. Zero

Example: A block slides down a rough inclined surface. The forces acting on the block are depicted below. The work done by the frictional force is:

Wf = |fk| |Δx| cos(180°) = -|fk| |Δx| < 0

Work done by the normal force: WN = |N| |Δx| cos(90°) = 0

Work done by weight: Wmg = |mg| |Δx| cos(θ ) > 0

0 < θ < 90°

N

mg

y

x

f

3

2. Work kinetic energy principle

Fdxxmamvmv

xxavv

)(22

)(2

12

21

22

1221

22

22

21

22 mvmv

W

2

2mvK Definition:

W=K2 - K1

Example: An 80-g arrow is fired from a bow whose string exerts an average force of 100 N on the arrow over a distance of 49 cm. What is the speed of the arrow as it leaves the bow?

m = 80 gF = 100 Nd = 49 cmv1= 0v2 - ? 2

022

2

1

mvK

K

FdW

2

22mv

Fd m

Fdv

22

smkg

mNv /35

1080

104910023

2

2

4

Example: Two blocks (m1=2m2) are pushed by identical forces, each starting at rest at the same start line. Which object has the greater kinetic energy when it reaches the same finish line? 

Same force, same distance Same work

Same change in kinetic energy

1. Box1               2. Box 23. They both have the

same kinetic energy

Example: A ball is dropped and hits the ground 50 m below. If the initial speed is 0 and we ignore air resistance, what is the speed of the ball as it hits the ground?

We can use kinematics or… the WKE theorem

2v = 2gh = 2(9.8 m / s )(50 m) = 31 m / s

Work done by gravity: mgh

0 221 mvmghKW

5

3.Gravitational potential energy. Conservation of energy

h

1

2

)(22

1212

12

21

22

UUKK

)hmg(hmghUWmvmv

WU Definition:

mghU g )( 12 hhmgmghW

mgF

g

g

constEUKUK 2211

0 EUK

For closed isolated system

Conservation of mechanical energy:

6

Example: A box of unknown mass and initial speed v0 = 10 m/s moves up a frictionless incline. How high does the box go before it begins sliding down?

mmghmv 002

021

2211 UKUK

m

sm

sm

g

vh 5

/102

/10

2 2

220

Only gravity does work (the normal

is perpendicular to the motion), so mechanical energy is conserved.

We can apply the same thing to any “incline”!

h

Turn-around point: where K = 0

E K UE K UE K U

v = 0

7

mghU initial h

final initial EE 2

2mv mgh

ghv 22

2mvK final

Example: A roller coaster starts out at the top of a hill of height h.

How fast is it going when it reaches the bottom?

Example: An object of unknown mass is projected with an initial speed, v0 = 10 m/s at an unknown angle above the horizontal. If air resistance could be neglected, what would be the speed of the object at height, h = 3.3 m above the starting point?

?

3.3

/100

v

mh

smv

smmsmsmghvv

mgmv

mghmv

/0.63.3/8.92/102

022

2220

20

2

8

Only weight of the pendulum is doing work; weight is a conservative force, so mechanical energy is conserved:

Lm

θ0

The angle on the other side is also θ0!

θ0

constUK

max

0

U

K

max

0

U

K

2

2mvK

mghU

min

max

U

K

Example: Pendulum (Conservation of energy)

9

Example: Pendulum (Conservation of energy)

The pendulum with a mass of 300 g is deviated from the equilibrium position B to the position A as shown below. Find the speed of the pendulum at the point B after the pendulum is released.

0 AA KmghU

A. Energy of the pendulum at the point A:

B. Energy of the pendulum at the point B:

0BU

02

02

mvmgh

UKUK BBAA

C. Conservation of energy:

2

2mvK B

smmsmghv /97.12.0/8.922 2

cmh 20

B

A

v

10

4.Hook’s Law (elastic forces)

FextFextFext

F

CkxU 221

2212

21

ifextsystemby kxkxWWU

kxF

Potential energy

11

Example: A box of mass m = 0.25 kg slides on a horizontal frictionless surface with an initial speed v = 10 m/s. a) How far will it compress the spring before coming to rest if k = 2500 N/m?

x

v = 10 m/sm = 0.25 kg k = 2500 N/m

Work-kinetic energy theorem: W = ΔKE

221 kxWext

m

mN

kgsmx 1.0

/2500

25.0/10

2212

21 mvkx

0221 mvK k

mvx

b) If the initial speed of box is doubled and its mass if halved, how far would the spring be compressed?

21

2

1

2

1

2 m

m

v

v

x

x

12

5. Conservative and nonconservative forces

•Forces such as gravity or the elastic force, for which the work does not depend on the path taken but only on the initial and final position, are called conservative forces

•For conservative forces work done on a closed path (a lop) is equal to zero

•Friction is a nonconservative force

213

A

B

Example: A block is moved from rest at point A to rest at point B.Which path requires the most work to be done on the object?

A) The table is leveled and friction is present.Path 1. Path 2. Path 3. All the same

B) The table is tilted and frictionless. Path 1. Path 2. Path 3. All the same

13

6. Power

1W=1J/s

t

dFP

vFvFP ||

Horsepower:1 hp = 746 WUnits:

Definition:t

WP

Example: You move 500 kg of bricks 1.0 m up. It takes 30 minutes by hand, or 10 minutes by lift. What is the power in each case?

Jm))(s

m.kg)((FdW 49000.189500

2

W.)s(

J

t

WP

handhand 72

min60min30

4900

Work is the same:

W.Pt

WP hand

liftlift 283

14

Example: Two elevators A and B carry each a load of mass m from the first floor to the third floor of a building at constant speeds, but A is twice as fast as B. 1) Compare work done by the cable tension (ie, the energy produced by the engine). 2) Compare power of two elevators.

T =mg

mg

Δx

t

WP

BA

AB

vv

tt

2

2

BA PP 2

BA WW xmgW 1)

2)

15