1. Work

cosFdW

[W] = N*m = JUnits:

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

Work and energy

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

gyF gxF

N

mgFg

y

x

f

Work done by the normal force:

WN = |N| |Δx| cos(90°) = 0

Work done by weight:

Wmg = mg|Δx| cos(θ ) > 0

Definition:

1

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

2

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

Work done by gravity: mgh

0 221 mvmghKW

3 smmsmghv /3150/8.922 2

3. Potential energy

forceveconservatiWU

8. Conservation of energy in mechanics

b) Elastic potential energy (spring):

mghU a) Gravitational potential energy:

2

2kxU kxF

mgF

01212 UUKK2211 UKUK

0 UK

4

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

5

mghU initial h

final initial EE

2

2mv mgh ghv 2

2

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

6

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)

7

4. Energy in the simple harmonic motion

kxF 221 kxU 2

212

21 kxmvE

tAv

tAx

sin

cos

Total mechanical energy is constant throughoscillation: conservation of energy!

U

x

E

–A A

U

K

22212

21

2221

22221

cos

sin

AmkAE

tkAU

tAmK

E

t

2mk

t

t

8

5. Damped Harmonic Motion

x(t)

tbvFd

makxbvF

tmbt eAeAtA

ttAtx2/

00)(

)'cos()()(

2

220 2

'

m

b

m

k

2 20

m

b

m

k

b – damping constant(Shows how fast oscillations decay)

Damping force is proportional to velocity:

Optional math:

9

6. Resonance

10