work & energy (usaha & energy)

Physics 211: Lecture 9, Pg 1

Physics 211: Lecture 9 and 10 combinedPhysics 211: Lecture 9 and 10 combined

Today’s AgendaToday’s Agenda

Work & Energy Discussion Definition

Dot Product Work of a constant force

Work/kinetic energy theorem Work of multiple constant forces Work done by variable forces

Spring 3-D generalization to all this Conservative forces and potential energy Comments

I’ll work out the Spring 2006 HE1 tonight from 7-9pm in Lincoln Hall TheaterYou can download it from the course web page


test this Thursday eveningtest this Thursday eveningwhere are you supposed to where are you supposed to

be?be?

Constant acceleration problems (including multiparticle) Centripetal acceleration Inertial reference frames and transformations Newton’s Laws of motion Newton’s Law of Gravity Friction: static and dynamic Free body diagrams, itemizing forces Tension in ropes (a.k.a. lines) Many body systems

Atwood’s machine, etc Springs

bring your I-cardbring your I-card


Today: show that net work equals change in Today: show that net work equals change in kinetic energykinetic energy

What is work? What is kinetic energy? Strategy:

Use Newton’s laws to derive result mathematically Interpret result (work, kinetic energy) All this comes from Newton + math No new assumptions A lot of logic…

Work is tied to forces Kinetic energy is tied to object (its mass, velocity)

Let equation talk to you


Definition of Work:Definition of Work:

Ingredients: Ingredients: Force (FF), displacement (rr)

Work, W, of a constant force FF

acting through a displacement rr

is:

W = FF rr = F rr cos = Fr rr

FF

rr

displace

ment

Fr

“Dot Product”


Definition of Work...Definition of Work...

Only the component of F along the displacement is doing work.

Example: cart on track

FF

rr

F cos

Hairdryer

W = FF rr = F rr cos = Fr rr


Aside: Dot Product (or Scalar Product)Aside: Dot Product (or Scalar Product)

Definition:

aa.bb = ab cos

= a[b cos ] = aba

= b[a cos ] = bab

Some properties:aabb = bbaaq(aabb) = (qbb)a a = bb(qaa) (q is a scalar)aa(b b + cc) = (aabb) + (aacc) (cc is a vector)

Units multiplyThe dot product of perpendicular vectors is 0 !!

aa

ab bb

aa

bb

ba


Aside: Examples of dot productsAside: Examples of dot products

Suppose

aa = 1 i i + 2 j j + 3 k k

bb = 4 i i - 5 j j + 6 k k

Then

aa . bb = 1x4 + 2x(-5) + 3x6 = 12aa . aa = 1x1 + 2x2 + 3x3 = 14bb . bb = 4x4 + (-5)x(-5) + 6x6 = 77

i i . ii = j j . j j = k k . k k = 1

i i . jj = j j . k k = k k . i i = 0x

y

z

ii

jj

kk

aa . bb = (1 i i + 2 j j + 3 k k )).((4 i i - 5 j j + 6 k k ))


Aside: Properties of dot productsAside: Properties of dot products

Magnitude:a2 = |a|2 = a . a

= (ax i i + ay jj) . (ax i i + ay jj)= ax

2(i i . ii) + ay 2(j j . jj) + 2ax ay (i i . jj)

= ax 2 + ay

2

Pythagorean Theorem!!

aa

ax

ay

ii

j j


Aside: Properties of dot productsAside: Properties of dot products

Components:aa = ax i i + ay j j + az k k = (ax , ay , az) = (aa . i i, aa . j j, aa . k k)

Derivatives:

Apply to velocity

So if v is constant (like for UCM):

ddt

ddt

ddt

( )a ba

b ab

ddt

vddt

ddt

ddt

2 2 ( )v vv

v vv

v a

ddt

v 2 2 0 v a


Back to the definition of Work:Back to the definition of Work:

Work, W, of a force FF acting

through a displacement rr is:

W = FF rrFF

rr

Skateboard


Lecture 9, Lecture 9, Act 1Act 1Work & EnergyWork & Energy

A box is pulled up a rough ( > 0) incline by a rope-pulley-weight arrangement as shown below.

How many forces are doing work on the box?

(a)(a) 2

(b)(b) 3

(c)(c) 4


Lecture 9, Lecture 9, Act 1Act 1SolutionSolution

Draw FBD of box:N

f

mg

T Consider direction of

motion of the box

v

Any force not perpendicularto the motion will do work:

N does no work (perp. to v)

T does positive work

f does negative work

mg does negative work

3 forcesdo work


Work: 1-D Example Work: 1-D Example (constant force)(constant force)

A force FF = 10 N pushes a box across a frictionlessfloor for a distance x x = 5 m.

xx

FF

Work done byby F F onon box : WF = FFxx = F x (since FF is parallel to xx)

WF = (10 N) x (5 m) = 50 Joules (J)Joules (J)


Units:Units:

N-m (Joule) Dyne-cm (erg)

= 10-7 J

BTU = 1054 J

calorie = 4.184 J

foot-lb = 1.356 J

eV = 1.6x10-19 J

cgs othermks

Force x Distance = Work

Newton x

[M][L] / [T]2

Meter = Joule

[L] [M][L]2 / [T]2


Work & Kinetic Energy:Work & Kinetic Energy:

A force FF = 10 N pushes a box across a frictionlessfloor for a distance x x = 5 m. The speed of the box is v1 before the push and v2 after the push.

xx

FFv1 v2

ii

m


Work & Kinetic Energy...Work & Kinetic Energy...

Since the force FF is constant, acceleration aa will be constant. We have shown that for constant a:

v22 - v1

2 = 2a(x2-x1) = 2ax. multiply by 1/2m: 1/2mv2

2 - 1/2mv12 = max

But F = ma 1/2mv22 - 1/2mv1

2 = Fx

xx

FFv1 v2

aa

ii

m


Work & Kinetic Energy...Work & Kinetic Energy...

So we find that 1/2mv2

2 - 1/2mv12 = Fx = WF

Define Kinetic Energy K: K = 1/2mv2

K2 - K1 = WF

WF = K (Work/kinetic energy theorem)(Work/kinetic energy theorem)

xx

FFaa

ii

m

v2v1


Work/Kinetic Energy Work/Kinetic Energy Theorem:Theorem:

{NetNet WorkWork done on object}

=

{changechange in kinetic energy kinetic energy of object}

KWnet

12 KK

21

22 mv

21

mv21

We’ll prove this same thing for a variable force in a bit.



Two blocks have masses m1 and m2, where m1 > m2. They are sliding on a frictionless floor and have the same kinetic energy when they encounter a long rough stretch (i.e. > 0) which slows them down to a stop. They experience same

(Which one will go farther before stopping?

(a)(a) m1 (b)(b) m2 (c)(c) they will go the same distance

m1

m2



The work-energy theorem says that for any object WWNETNET = = KK In this example the only force that does work is ffriction (since both N and mg are perpendicular to the block’s

motion).

mf

N

mg


Lecture 9, Lecture 9, Act 2Act 2 SolutionSolution

The work-energy theorem says that for any object WWNETNET = = KK In this example the only force that does work is ffriction (since

both N and mg are perpendicular to the blocks motion). The net work done to stop the box is - fD = -mgD.

m

D

This work “removes” the kinetic energy that the box had: WNET = K2 - K1 = 0 - K1


Lecture 9, Lecture 9, Act 2Act 2 SolutionSolution

The net work done to stop a box is - fD = -mgD. This work “removes” the kinetic energy that the box had: WNET = K2 - K1 = 0 - K1

This is the same for both boxes (same starting kinetic energy).

m2gD2m1gD1 m2D2m1D1

m1

D1

m2

D2

Since m1 > m2 we can see that D2 > D1


A simple application:A simple application:Work done by gravity on a falling objectWork done by gravity on a falling object

What is the speed of an object after falling a distance H, assuming it starts at rest?

Wg = FF r r = mg rr cos(0) = mgH

Wg = mgH

Work/Kinetic Energy Theorem:Work/Kinetic Energy Theorem:

Wg = mgH = 1/2mv2

rrmg g

H

j j

v0 = 0

v v gH 2


Work done by gravity:Work done by gravity:

Wg = FF rr = mg rr cos = -mg y

(remember y = yf - yi)

Wg = -mg y

Depends only on y !

not x

j j

m

rrmg g

y

m


Work done by gravity taking a funny path...Work done by gravity taking a funny path...

Depends only on y,

not on path taken!

This is a very artificial example

Be careful of very artificial examples!

m

mg g

y j j

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

r

= F r

= F yrr11rr22

rr33

rrnn

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

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

Wg = -mg y


What about multiple forces?What about multiple forces?

Suppose FFNET = FF1 + FF2 and the

displacement is rr.

The work done by each force is:

W1 = FF1 r r W2 = FF2 rr

WTOT = W1 + W2

= FF1 r r + FF2 rr

= (FF1 + FF2 ) rr

WTOT = FFTOT rr It’s the totaltotal force that matters!!

FFNET

rrFF1

FF2


Comments:Comments:

Time interval not relevant Run up the stairs quickly or slowly...same W

Since W = FF rr

No work is done if: FF = 0 or rr = 0 or = 90o


Comments...Comments...

W = FF rr

No work done if = 90o.

No work done by TT.

No work done by N.

TT

v v

vvNN


Lecture 10, Lecture 10, Act 1Act 1Falling ObjectsFalling 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

Fallingobjects



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)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. Remember integration is just sophisticated summing

F

x

Wg

x

F(x)

x1 x2 dx 2

1

x

x

dx)x(FW


Work/Kinetic Energy Theorem for a Work/Kinetic Energy Theorem for a

VariableVariable Force Force

2

1

x

x

W F dx

dx2

1

x

x dtm dv

2

1

v

v

m v dv

dtmma F dv

dvdx

dxv2

1

x

x

m

dxdxdv dv dv

dxv (chain rule)

dt=

dt=

ΔKEm21

m21

)(21

m v22 v1

2 v22 v1

2

energykineticmvmvrdFWr

r

212

1222

12

1


1-D Variable Force Example: Spring1-D 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

the mass


Spring...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...Spring...

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

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.

Spring

In this example it is a negative number. The spring does negative work on the mass



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



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

k

mvx 1

11In the case of x1



x2v2

k

mvx

m2

m2

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

k

2mv

k

2mv2x 1

11

12

12x2x


Problem: Spring pulls on mass.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


Problem: Spring pulls on mass.Problem: Spring pulls on mass. First find the net work done on the mass during the motion from x = d to x = 0 (only due to the spring):


d

relaxed position

vr

m

m

ii

2222

1

2

2skd

2

1d0k

2

1xxk

2

1W


Problem: Spring pulls on mass.Problem: Spring pulls on mass. Now find the change in kinetic energy of the mass:


d

relaxed position

vr

m

m

ii

2r

21

22 mv

21

mv21

mv21

ΔK


Problem: Spring pulls on mass.Problem: Spring pulls on mass. Now use work kinetic-energy theorem: Wnet = WS = K.


d

relaxed position

vr

m

m

ii

1

22kd 2

rmv21

mk

dv r


Work by variable force in 3-D:Work by variable force in 3-D:

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


Work/Kinetic Energy Theorem for a Work/Kinetic Energy Theorem for a

VariableVariable Force in 3D Force in 3D

energykineticmvmvrdFWr

r

212

1222

12

1

Sum up F.dr along path

That’s the work integral

That equals change in KE

For conservative forces, the work is path independent and depends only on starting point and end point


Work by variable force in 3-D:Work by variable force in 3-D:Newton’s Gravitational ForceNewton’s Gravitational Force

Work dWg done on an object by gravity in a displacement drdr isgiven by:

dWg = FFg.drdr = (-GMm / R2 rr).(dR rr + Rd)

dWg = (-GMm / R2) dR (since rr. = 0, rr.r r = 1)

^̂ ^̂^̂

rr^̂

^̂drdrRd

dR

R

FFg m

M

d

^̂ ^̂ ^̂ ^̂



Integrate dWg to find the total work done by gravity in a “big”displacement:

Wg = dWg = (-GMm / R2) dR = GMm (1/R2 - 1/R1)

FFg(R1)

R1

R2

FFg(R2)

R1

R2

R1

R2

m

M



Work done depends only on R1 and R2, not on the path takennot on the path taken.

R1

R2

m

M

12

g R

1

R

1GMmW


Newton’s Gravitational ForceNewton’s Gravitational ForceNear the Earth’s Surface:Near the Earth’s Surface:

Suppose R1 = RE and R2 = RE + y

but we have learned that

So: Wg = -mgy

y

R

GMm

RyR

RyRGMm

RR

RRGMmW

2

EEE

EE

21

12g

GM

Rg

E2

RE+ y

M

m

RE

1 2

1 1gW GMm

R R


Conservative 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: 2

1

2

2sxxk

2

1W

ymgWg

12

g R

1

R

1GMmW


Conservative Forces: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 0.

Potential energy change from one point to another does not depend on path


Recap of today’s lectureRecap of today’s lecture

Work & Energy(Text: 6-1 and 7-4)

Dot Product(Text: 6-2)

Work of a constant force (Text: 7-1 and 7-2)

Work/kinetic energy theorem (Text: 6-1) Work done by variable force (Text: 6-1)

Spring Work done by variable force in 3-D (Text: 6-1)

Newton’s gravitational force (Text: 11-2) Conservative Forces

(Text: 6-4)

Look at textbook problems Look at textbook problems Chapter 6: # 1, #20, 23, 24, 25, 26, 34#20, 23, 24, 25, 26, 34 45

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