Welcome back to PHY 183

Meaning of the picture ? PE KE

3.1 Work3.2 Energy3.3 Conservative and nonconservative forces3.4 Power, energy and momentum conservation3.5 Linear momentum3.6 Collisions3.7 Motion in a gravitational potential

CHAPTER 3 WORK AND ENERGY

Part 1

Work• Work of constant Force• Work done by sum of constant Forces• Work done by Variable Force

Review: Work of constant Force

Work ( W) of a constant force F F acting through a displacement r r is:

W = FF r r = F r cos() = Fr r (N.m=J)r (N.m=J)

FF

rr

displace

ment

Fr

Learning check

Find work done by tension T and by weigh

Find work done by Normal force NN.

What is your conclusion from these ??

WT= 0 and WW = ?

WN= 0

vv

NN

TT

v v

Test: Work done by gravity

WG =FF rr =mg. r.r.cos

= -mg y (remember y = yf - yi)

WG = -mg y

Depends only on y !

Let mass m move on the path rr by gravity only

Compute work done ??

j j

m

rrmg g

y

m

Work done by sum of constant Forces

Suppose FFNET = FF1 + FF2 + FF3 …+ FFn and the displacement is rr.

The work done by each force is:

W1 = FF1 r r W2 = FF2 r

…. Wn = FFn r

WNET = W1 + W2 + W3 …+ Wn

= FF1 r r + FF2 r…+ r…+ FFn rr

WNET = FFNET rr

Net work doneFFTOT

rrFF1

FF2

FF3

FFn

Test: Work done by gravity

Let mass m move on the path (rr11 + rr 2+ . . .+ rrnn) by gravity

Compute work done on total path:

WNET = W1 + W2 + . . .+ Wn

Note: Force = weight= F = mg

= FF rr 1+ FF rr2 + . . . + FF rrn

= FF (rr11 + rr 2+ . . .+ rrnn)

= F r

= F y

Depends only on y,

not on path taken!

r

m

mg g

y j j

rr11rr22

rr33

rrnn

Work done by Variable Force:

• When the force was constant, we wrote W = F x– area under F vs. x plot:

• For variable force, we find the areaby integrating:– dW = F(x) dx.

F

x

Wg

x

2

1

x

x

dx)x(FW

F(x)

x1 x2 dx

Work of variable force Example: Spring

• For a spring we know that Fx = -kx.

F(x) x2

x

x1

-kxrelaxed position

F = - k x1

F = - k x2

Problem

• Compute work done by the spring Ws during a displacement from x1 to x2 (Note: the F(x) vs x plot between x1 and x2).

Ws

F(x) x2

x

x1

-kxrelaxed position

Solution

2

1

2

2s

x

x

2

x

x

x

xs

xxk2

1W

kx2

1

dxkx

dxxFW

2

1

2

1

2

1

)(

)(

F(x) x2

Ws

x

x1

-kx

Work by variable force in 3D

Work dWF of a force F F acting

through an infinitesimal

displacement r r is:

dW = FF.rr The work of a big displacement through a variable

force will be the integral of a set of infinitesimal displacements:

WTOT = FF.rr

FF

rr

Part 2

EnergyKinetic energy KE: moving energy

Potential energy PE: Tendency for work Total energy: TE=KE + PE

Kinetic Energy Theorem

{NetNet WorkWork done on object}=={changechange in kinetic energy kinetic energy of object}

WF = K = 1/2mv22 - 1/2mv1

2

xx

FFv1 v2

m WF = Fx

Prove: Kinetic Energy Theorem for a variable Force

2

1

x

x

W dtmma

ΔKEm21

m21

)(21

m

FF dx

dx2

1

x

x dtm

dv

dx

dv

2

1

v

v

m v dv

v22 v1

2 v22 v1

2

dv

dxv

dx

dxdv dv dvdx

v (chain rule)

2

1

v

v

m

dt=

dt=

Test: Falling Objects

• Three objects of mass m begin at height h with velocity 0. One falls straight down, one slides down a frictionless inclined plane, and one swings on the end of a pendulum. What is the relationship between their velocities when they have fallen to height 0?

v=0

vi

H

v=0

vp

v=0

vf

Free Fall Frictionless incline Pendulum

(a)(a) Vf > Vi > Vp (b) (b) Vf > Vp > Vi (c) (c) Vf = Vp = Vi

Solution

v=0

vi

H

v=0

vp

v=0

vf

Free Fall Frictionless incline Pendulum

Only gravity will do work: Wg = mgH = 1/2 mv22 - 1/2 mv1

2 = 1/2 mv22

gH2vvv pif does not depend on path !!

Learning check

• A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed position while momentarily coming to rest.– If the initial speed of the box were doubled and its mass were halved, how far x2 would the spring compress ?

x1

(a)(a) (b) (b) (c)(c)12

xx 12

x2x 12

x2x

Solution• Again, use the fact that WNET = K.

In this case, WNET = WSPRING = -1/2 kx2

and K = -1/2 mv2

so kx2 = mv2

k

mvx 1

11In the case of x1

k

mvx So if v2 = 2v1 and m2 = m1/2

k

2mv

k

2mv2x 1

11

12

12x2x

x1v1

m1

m1

Problem: Spring pulls on mass.

• A spring (constant k) is stretched a distance d, and a mass m is hooked to its end. The mass is released (from rest). What is the speed of the mass when it returns to the relaxed position if it slides without friction?

relaxed position

stretched position (at rest)

dafter release

back at relaxed position

vr

v

m

m

m

m

Step-1• First find the net work done on the mass during the motion from x = d to x = 0 (only due to the spring):

stretched position (at rest)

d

relaxed position

vr

m

m

ii

2222

1

2

2skd

2

1d0k

2

1xxk

2

1W

Step-2

• Now find the change in kinetic energy of the mass:

stretched position (at rest)

d

relaxed position

vr

m

m

ii

2r

21

22 mv

21

mv21

mv21

ΔK

Step-3

• Now use work kinetic-energy theorem: Wnet = WS = K.

stretched position (at rest)

d

relaxed position

vr

m

m

ii

1

22kd 2

rmv21

m

kdvr

Step-4

• Now suppose there is a coefficient of friction between the block and the floor• The total work done on the block is now the sum of the work done by the spring WS (same as before) and the work done by friction Wf.

Wf = ff.ΔrΔr = - mg d

stretched position (at rest)

d

relaxed position

vr

m

m

ii

f = mg

rr

Step-4

• Again use Wnet = WS + Wf = K

Wf = -mg d

stretched position (at rest)

d

relaxed position

vr

m

m

ii

f = mg

rr

W kdS 1

22 2

rmv21

K

2r

2 mv21

mgdkd21 gd2d

mk

v 2r μ

Part 3

Conservative and nonconservative forces

Conservative Forces:

• In general, if the work done does not depend on the path taken (only depends the initial and final distances between objects), the force involved is said to be conservativeconservative.

Gravity is a conservative force:

• Gravity near the Earth’s surface:

• A spring produces a conservative force:

21

22s xxk

2

1W

ymgWg

12

g R

1

R

1GMmW

Result of conservative Forces

• We have seen that the work done by a conservative force does not depend on the path taken.

W1

W2

W1

W2

W1 = W2

WNET = W1 - W2

= W1 - W1 = 0

• Therefore the work done in a closed path is zero.

Potential Energy

• For any conservative force F we can define a potential energy function U in the following way:

– The work done by a

conservative force is equal

and opposite to the change

in the potential energy function.

This can be written as:

W = FF.dr r = -U = U1 – U2

U = U2 - U1 = -W = - FF.drrr1

r2 r1

r2 U2

U1

Test: Gravitational Potential Energy

• We have seen that the work done by gravity near the Earth’s surface when an object of mass m is lifted a distance y is Wg = -mg y

• Compute the change in potential energy of this object ??

U = -Wg = mg y

ym

Wg = -mg y

jj

Gravitational Potential Energy

• So we see that the change in U near the Earth’s surface is: U = -Wg = mg y = mg(y2 -y1).

So U = mg y + U0 where U0 is an arbitrary constantarbitrary constant.

Having an arbitrary constant U0 is equivalent to saying that we can choose the y location where U = 0 to be anywhere we want to.

y1

m

Wg = -mg y

jj y2