Physics 2011

Physics 2011

Chapter 6: Work and Kinetic Energy

Work

The Physics of Work

• By strict definition, in order for work to be performed, a Net Force must be applied to a body, resulting in the Displacement of that body.

Work = Force * Displacement = Newtons * Meters = Joules (Energy)

Calculating Work from Vectors• Consider the Idiot pushing his girlfriend’s car in

one direction while she steers in another:

• The useful work is:

• Thus the Scalar, Work, is a DOT PRODUCT:

Work has a Sign• Work is calculated by finding the component

of Force acting along the line of Displacement, but they may be in opposite directions.

• ALSO, Work is W and Weight is w …..OK?

Work is ENERGY

• Work is the product of a Net Force and an accompanying displacement

• A body under the influence of a Net Force is accelerating (F = ma)

• An accelerating body is said to have increasing Kinetic Energy

Kinetic Energy

• A body with Mass, m, moving at velocity, v, has some ability to perform Work(For example, a bowling ball rolling down the alley can knock over pins)

• This ability of a moving body to do work (Work is Energy) is quantified as:

Kinetic Energy, K = ½ mv2 (Joules)

Work-Energy

• Positive Work on a Body INCREASES its Kinetic Energy

• Negative Work on a Body DECREASES its Kinetic Energy

• A body that gains K must increase in speed and a body that loses K must decrease in speed.

gotta have POWER!!!!

Power is the RATE of Work:

i.e. Power is the change in work over some unit of time

P = ΔW / Δt (Average Power)

P = dW/dt (Instantaneous Power)

Power is Joules/Seconds or Watts

Review: Sum of Constant Forces

Suppose FFNET = FF1 + FF2 and thedisplacement is SS.

The work done by each force is:

W1 = FF1 r r W2 = FF2 r

FFTOTrrFF1

FF2 WNET = W1 + W2

= FF1 r r + FF2 rr = (FF1 + FF2 ) rr

WNET = FFNET rr

Review: Constant Force... W = FF rr

• No work done if = 90o.

– No work done by TT.

– No work done by NN.

vvNN

TT

v v

Work/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

Work done by gravity:

• Wg = FF rr = mg rr cos = -mg y

(remember y = yf - yi)

Wg = -mg y

Depends only on y !

j j

m

rrmg g

y

m

Work done by gravity...•

Depends only on y, not on path taken!

m

mg g

y j j

W NET = W1 + W2 + . . .+ Wn

r

= F r= F y

rr11rr22

rr33

rrnn

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

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

Wg = -mg y

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?

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

v=0

vi

H

v=0

vp

v=0

vf

Free Fall Frictionless incline Pendulum

Solution

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

2 = 1/2 mv22

gH2vvv pif does not depend on path !!

v = 0

vi

H

v = 0

vp

v = 0

vf

Free Fall Frictionless incline Pendulum

Work done by Variable Force: (1D)

• 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

xdx)x(FW

F(x)

x1 x2 dx

Work/Kinetic Energy Theorem for a Variable Force

2

1

x

xW dt

mma

ΔKEm21m

21)(

21m

FF dx

dx2

1

x

x dtm dv

dxdv

2

1

v

vm v dv

v22 v1

2 v22 v1

2

dv

dxv

dxdxdv dv dv

dxv (chain rule)

2

1

v

vm

dt = dt =

1-D Variable Force Example: Spring

• For a spring, Hooke’s Law states: Fx = -kx.

F(x) x2

x

x1

-kxrelaxed position

F = - k x1

F = - k x2

Spring...• The work done by the spring Ws during a displacement from x1 to x2 is the area under the F(x) vs x plot between

x1 and x2.

Ws

F(x) x2

x

x1

-kxrelaxed position

Spring...

21

22s

x

x

2

x

x

x

xs

xxk21W

kx21

dxkx

dxxFW

2

1

2

1

2

1

)(

)(F(x) x2

Ws

x

x1

-kx

• The work done by the spring Ws during a displacement from x1 to x2 is the area under the F(x) vs x plot between x1 and x2.

Work & Energy• 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 ?

x

(a)(a) (b) (b) (c)(c)12 xx 12 x2x 12 x2x

Lecture 10, Act 2Solution

• Again, use the fact that WNET = K.

x1v1

so kx2 = mv2

m1

m1

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

and K = -1/2 mv2

km

vx 111 In the case of x1

Lecture 10, Act 2Solution

x2v2

kmvx

m2

m2

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

k2m

vk

2mv2x 1

11

12

12 x2x