Physics II, Pg 1 Physics II Today’s Agenda l Work & Energy. çDiscussion. çDefinition. l Scalar...

Physics II, Pg 1

Physics IIPhysics II

Today’s AgendaToday’s Agenda Work & Energy.

Discussion.Definition.

Scalar Product. Work of a constant force.

Work kinetic-energy theorem. Work of a sum of constant forces. Work for a sum of displacements with constant force. Comments. Look at textbook problems Chp7 -7,11,12,17,20,23,24,27,29,36Look at textbook problems Chp7 -7,11,12,17,20,23,24,27,29,36

Physics II, Pg 2

Work & EnergyWork & Energy

One of the most important concepts in physics.Alternative approach to mechanics.

Many applications beyond mechanics.Thermodynamics (movement of heat).Quantum mechanics...

Very useful tools.You will learn new (sometimes much easier) ways to

solve problems.

See text: 1-1 and 7-1

Physics II, Pg 3

Forms of EnergyForms of Energy

KineticKinetic: Energy of motion.A car on the highway has kinetic energy.

We have to remove this energy to stop it.The breaks of a car get HOT !This is an example of turning one form of energy into

another. (More about this soon)...


Physics II, Pg 4

Forms of EnergyForms of Energy

PotentialPotential: Stored, “potentially” ready to use.Gravitational.

» Hydro-electric dams etc...Electromagnetic

» Atomic (springs, chemical...)Nuclear

» Sun, power stations, bombs...


Physics II, Pg 5

Mass = Energy Mass = Energy (but not in Physics II)(but not in Physics II)

Particle Physics:

+ 5,000,000,000 V

e-

- 5,000,000,000 V

e+(a)

(b)

(c)

E = 1010 eV

M E = MC2

( poof ! )

Physics II, Pg 6

Energy ConservationEnergy Conservation Energy cannot be destroyed or created.

Just changed from one form to another.

We say energy is conservedenergy is conserved !True for any isolated system. i.e when we put on the brakes, the kinetic energy of the car is turned into heat using friction in the brakes. The total energy of the “car-breaks-road-atmosphere” system is the same.The energy of the car “alone” is not conserved...

» It is reduced by the braking.

Doing “workwork” on a system will change it’s “energyenergy”...

See text: 7-1

Physics II, Pg 7

Definition of Work:Definition of Work:

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

Work, W, of a constant force FF

acting through a displacement SS

is:

W = FF..SS = FScos() = FS S

FF

SS

displace

ment

FS

“Dot Product”


Physics II, Pg 8

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

Definition:

aa.bb = abcos()

= a[bcos()] = aba

= b[acos()] = bab

Some properties:aa.bb = bb.aaq(aa.bb) = (qbb).a a = bb.(qaa) (q is a scalar)aa.(b b + cc) = (aa.bb) + (aa.cc) (cc is a vector)

The dot product of perpendicular vectors is 0 !!

aa

ab

bb

ba

See text: 7-2

Physics II, Pg 9

Aside: Examples of dot productsAside: Examples of dot products

Suppose Then

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

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

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 = 0

See text: 7-2

Physics II, Pg 10

Aside: Properties of dot productsAside: Properties of dot products

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

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

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

= ax 2 + ay

2

Pythagoras Theorem !!

aa

ax

ay

ii

j j


Physics II, Pg 11

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

See text: 7-2

since a and v are perpendicular

Physics II, Pg 12

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

Work, W, of a force F F acting

through a displacement S S is:

W = FF..SSFF

SS

See text: 7-2

Physics II, Pg 13

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

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

xx

FF

Work done byby F F onon box :

WF = FF.xx = F x (since FF is parallel to xx)

WF = (50 N)x(5m) = 50 N-m.


See example 7.1

Physics II, Pg 14

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

See text: 7-1

Physics II, Pg 15

Work & Kinetic Energy:Work & Kinetic Energy:

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

xx

FFv1 v2

ii

m


Physics II, Pg 16

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

Since the force FF is constant, acceleration a a will be constant. We have shown that for constant a:v2

2 - v12 = 2a(x2-x1 ) = 2ax.

multiply by 1/2m: 1/2mv22 - 1/2mv1

2 = ma x

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

2 = Fx

xx

FFv1 v2

aa

ii

m

See text: 7-1, 7-2, 7-5

Physics II, Pg 17

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

So we find that1/2mv2

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

FFv1 v2

aa

ii

m

Physics II, Pg 18

Work Kinetic-Energy Theorem:Work Kinetic-Energy Theorem:

{NetNet WorkWork done on object}

=

{changechange in kinetic energy kinetic energy of object}

This is true in general:

K1

K2

FFnet

dSdS

W K K K mv mvnet 2 1 22

121

2

1

2

See text: 7-3, 7-4, 7-5

Physics II, Pg 19

Work done by Variable Force: (1D)Work done by Variable Force: (1D)

When the force was constant, we wrote W = Fxarea under F vs x plot:

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

x

Wg

x

W F x dxx

x

( )

1

2

F(x)

x1 x2 dx

See text: 7-3

Physics II, Pg 20

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.S S = mgScos(0) = mgH

Wg = mgH


Wg = mgH = 1/2mv2

SSmg g

H

j j

v0 = 0

v v gH 2

See text: 7-3, 7-4, 7-5

Physics II, Pg 21

What about a sum of forces?What about a sum of forces?

Suppose FFTOT = FF1 + FF2 and the

displacement is SS.

The work done by each force is:

W1 = FF1 ..S S W2 = FF2

..SS

WTOT = W1 + W2

= FF1 ..S S + FF2

..SS

= (FF1 + FF2 )..SS

WTOT = FFTOT ..SS It’s the totaltotal force that mattters !!

FFTOT

SSFF1

FF2

See text: 7-3, 7-4, 7-5

Physics II, Pg 22

Work by sum of displacementsWork by sum of displacementswith constant force.with constant force.

W = W1 + W2

= FF.SS1 + FF.SS2

= FF.( SS1 + SS2 )

W = FF.SS

Work depends onlyWork depends onlyon total displacement,on total displacement,

not on the “path”.not on the “path”.

SS1

SS2

SS

F F

See text: 7-3, 7-4, 7-5

Physics II, Pg 23

Work by sum of displacementsWork by sum of displacementswith constant force.with constant force.

SS

FFSS1

SS2

SSn

SS3

W = W1 + W2 + . . .+ Wn

= FF.SS1 + FF.SS2 + . . . + FF.SSn

= FF.( SS1 + SS2 + . . .+ SSn )

W = FF.SS

Same result as simple case.Same result as simple case.

See text: 7-3, 7-4, 7-5

Physics II, Pg 24

Comments:Comments:

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

Since W = FF.SS

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

See text: 7-3, 7-4, 7-5

Physics II, Pg 25

Comments...Comments...W = FF.SS No work done if = 90o.

No work done by TT.

No work done by N.

TT

v v

vvNN

See text: 7-3, 7-4, 7-5

Physics II, Pg 26

Recap of today’s lectureRecap of today’s lecture

Work & Energy.Discussion.Definition.

Scalar Product. Work of a constant force.

Work kinetic-energy theorem. Properties (units, time independence etc). Work of a combination of forces. Comments.

Look at textbook problems Serway Chp7 -Look at textbook problems Serway Chp7 -7,11,12,17,20,23,24,27,29,367,11,12,17,20,23,24,27,29,36

Physics II, Pg 27

Physics II: Lecture Physics II: Lecture

Todays AgendaTodays Agenda

Review of Work. Last lecture Work done by gravity near the earths surface. Examples:

pendulum, inclined plane, free-fall. Work done by variable force.

Spring Problem involving spring & friction- Serway 7- 37,39,40,46,51

Physics II, Pg 28


{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

See text: 7-3, 7-4, 7-5

Physics II, Pg 29

Work done by gravity:Work done by gravity:

Wg = FF.SS11 = mgS11cos(1)

= -mgS11cos(1)

= -mgy

Wg = -mgy

Depends only on y !m

SS11

mg g

y

1

1 j j

See text: 7-1

Physics II, Pg 30

Work done by gravity...Work done by gravity...

Wg = (FF.ss) = FF.SSTOT

Wg = -mgy

Depends only on y !m

mg g

y j j

See text: 7-1

See example 7-2 (easy) and example 7-9 (harder)

Physics II, Pg 31

Example: Falling ObjectsExample: Falling Objects

v=0

vf

H

v=0

vf

v=0

vf

Free Fall Frictionless incline Pendulum

See text: 7-1

Physics II, Pg 32

Example: Falling Objects...Example: Falling Objects... W = FF.S = S = FScos()

No work done if = 90o.

No work done by TT.Only mgg does work !

No work done by NN.Only mgg does work !

mggNN v v

TT

v v mg


Physics II, Pg 33

Example: Falling ObjectsExample: Falling Objects

v=0

vf

H

v=0

vf

v=0

vf

Free Fall Frictionless incline Pendulum

Only gravity will do work: Wg = mgH = 1/2 mvf2

v gHf 2 does not depend on path !!


Physics II, Pg 34

Lifting a book with your hand:Lifting a book with your hand:What is the total work done on the book ??What is the total work done on the book ??

First calculate the work done by gravity:

Wg = mgg..S S = -mgS

Now find the work done bythe hand:

WHAND = FFHAND.S S = FHAND S

mgg

SS FFHAND

vv = const

aa = 0

See text: 7-1 to 7-5

Physics II, Pg 35

Example: Lifting a book...Example: Lifting a book...

Wg = -mgS

WHAND = FHAND S

WTOT = WHAND + Wg

= FHAND S -mgS

= (FHAND -mg)S

= 0 since aa = 0

So WTOT = 0 !!

mgg

SS FFHAND

vv = const

aa = 0


Physics II, Pg 36

Example: Lifting a book...Example: Lifting a book...

Work Kinetic-Energy Theorem says: WF = K

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

In this case, vv is constant so K = 0and so WF must be 0, as we found.

mgg

SS FFHAND

vv = const

aa = 0

See text: 7-5

Physics II, Pg 37

Work done by Variable Force: (1D)Work done by Variable Force: (1D)

When the force was constant, we wrote W = Fxarea under F vs x plot:

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

x

Wg

x

W F x dxx

x

( )

1

2

F(x)

x1 x2 dx

See text: 7-3

Physics II, Pg 38

1-D Variable Force Example: Spring1-D Variable Force Example: Spring

For a spring we know that Fx = -kx.

F(x) x2

x

x1

-kxequilibrium

F = - k x1

F = - k x2

See text: 7-3

Physics II, Pg 39

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

-kxequilibrium

See text: 7-3

Physics II, Pg 40

Spring...Spring...

W F x dx

kx dx

kx

k x x

sx

x

x

x

x

x

( )

( )

1

2

1

2

1

21

2

1

2

2

22

12

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.

See text: 7-3

Physics II, Pg 41

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 equilibrium position if it slides without friction?

equilibrium position

stretched position (at rest)

dafter release

back at equilibrium position

vE

v

m

m

m

m

See text: 7-3

Physics II, Pg 42

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


ve

m

m

ii

W k x x k d kds 1

2

1

20

1

222

12 2 2 2

See text: 7-3

Physics II, Pg 43

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


d


ve

m

m

ii

K mv mv mve 1

2

1

2

1

222

12 2

See text: 7-3

Physics II, Pg 44

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


d


ve

m

m

ii

1

22kd 1

22mve

v dk

me

See text: 7-3

Physics II, Pg 45

Problem: Spring pulls on mass.Problem: Spring pulls on mass. 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.SS = - mg d


d


ve

m

m

ii

f = mg

SS

See text: 7-3

Physics II, Pg 46

Problem: Spring pulls on mass.Problem: Spring pulls on mass. Again use Wnet = WS + Wf = K

Wf = - mg d


d


ve

m

m

ii

f = mg

SS

W kdS 1

22 K mve

1

22

1

2

1

22 2kd mgd mve v

k

md gde 2 2

See text: 7-3

Physics II, Pg 47


Review Work done by gravity near the earths surface. Examples:

pendulum, inclined plane, free-fall. Work done by variable force.

Spring Problems involving spring & friction. Serway 7- 37,39,40,46,51

Physics II, Pg 48

Physics II, Pg 49

Physics II: Lecture Physics II: Lecture


Review. Work done by variable force in 3-D.

Newtons gravitational force. Conservative Forces & Potential energy. Conservation of “Total Mechanical Energy”

Example: Pendulum. Nonconservative force

friction General work-energy theorem Example problem. Serway Chp 8- 1,2,7,9,11,12,14,20,22,25,27,32,35

Physics II, Pg 50

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

Work dWF of a force F F acting

through an infinitesmal

displacement dS dS is:

dW = FF.dSdS

The work of a big displacement through a variable force will be the integral of a set of infinitesmal displacements:

WTOT = FF.dSdS

FF

dSdS

See text: 7-4

Physics II, Pg 51

Work by variable force in 3-D:Work by variable force in 3-D:Newtons Gravitational ForceNewtons Gravitational Force

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

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

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

^̂ ^̂^̂

rr^̂

^̂dsdsRd

dR

R

FFg m

M

d

^̂ ^̂

See text: 7-4, 8-1

Looking at sections 9-2 and 9-4 in the text might help.

Physics II, Pg 52


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


See text: 7-4, 8-1

Physics II, Pg 53


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

R1

R2

W GMmR Rg

1 1

1 2

m

M

See text: 7-4, 8-1


Physics II, Pg 54

Newtons Gravitational ForceNewtons Gravitational ForceNear the Earths Surface:Near the Earths Surface:

Suppose R1 = RE and R2 = RE + y

but we have learned that

So: Wg = -mgy

W GMmR R

R RGMm

R y R

R y Rm

GM

Ryg

E E

E E E

2 1

1 22

GM

Rg

E2

RE+ y

M

m

RE

See text: 7-4

See example 7-9

Physics II, Pg 55

Conservative Forces:Conservative Forces:

We have seen that the work done by gravity does not depend on the path taken, only on the change in separation between the two objects.

W GMmR Rg

1 1

1 2

R1

R2

M

m

hm

Wg = -mgh

See text: 8-1

See example 7-9

Physics II, Pg 56

Conservative Forces:Conservative Forces:

In general, if the work done does not depend on the path taken, the force involved is said to be conservativeconservative.

Gravity is a conservative force :

A spring produces a conservative force:

Any constant force is also conservative:

W GMmR Rg

1 1

1 2

W k x xs 1

2 22

12

SS

FF


Physics II, Pg 57

Potential EnergyPotential 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 equals (-) the change in the potential energy function.

This can be written as:

W = FF.dS S = - U

U = U2 - U1 = - W = - FF.dSSS1

S2

S1

S2 U2

U1

See text: 8-2

Physics II, Pg 58

Gravitational Potential EnergyGravitational Potential Energy

We have seen that the work done by gravity near the earths surface when an object of mass m is lifted a distance y is Wg = -mgy.

The change if potential energy of this object is therefore:

U = - Wg = mgy.

ym

Wg = -mgy

jj

See text 8-2

See example 8-1

Physics II, Pg 59

Gravitational Potential EnergyGravitational Potential Energy

So we see that the change in U near the earths surface is: U = - Wg = mgy = mg(y2 -y1).

So U = mgy + 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 = -mgy

jj y2

See text: 8-2

See example 8-1

Physics II, Pg 60

Potential Energy Recap:Potential Energy Recap:

For any conservative force we can define a potential energy function U such that:

The potential energy function U is always defined onlyup to an additive constant.

You can choose the location where U = 0 to be anywhere convenient.

U = U2 - U1 = - W = - FF.dSSS1

S2


Physics II, Pg 61

Conservative Forces & Potential Energies Conservative Forces & Potential Energies (stuff you should know):(stuff you should know):

Force Force

FF

WorkWork

W(1-2)

Change in P.EChange in P.E

U = U2 - U1

P.E. functionP.E. function

U

FFg = -mg j j

FFg = r r

Fs = -kx

^̂

^̂

GMmR R

1 1

2 1

1

2 22

12k x x

-mg(y2-y1) mg(y2-y1)

GMmR R

1 1

2 1

1

2 22

12k x x

GMm

R2

mgy + C

GMm

RC

1

22kx C

Physics II, Pg 62

Conservation of EnergyConservation of Energy

If only conservative forces are present, the total energy If only conservative forces are present, the total energy (sum of potential and kinetic energies) of a system(sum of potential and kinetic energies) of a system is is conserved (i.e. constant).conserved (i.e. constant).

E = K + U is constantconstant !!!!!!

Both K and U can change as long as E = K + U is constant.

See text: 8-3

Physics II, Pg 63

Example: The simple pendulum.Example: The simple pendulum.

Suppose we release a bob or mass m from rest a distance h1 above it’s lowest possible point.What is the maximum speed of the bob and where

does this happen ?To what height h2 does it rise on the other side ?

v

h1 h2

m

See text: 8-3

See example A Pendulum

Physics II, Pg 64


Energy is conserved since gravity is a conservative force (E = K + U is constant)

Choose y = 0 at the original position of the bob, and U = 0 at y = 0 (arbitrary choice).

E = 1/2mv2 + mgy.

v

h1 h2

y

y=0

See example , A Pendulum

See text: 8-3

Physics II, Pg 65


E = 1/2mv2 + mgy.Initially, y = 0 and v = 0, so E = 0.Since E = 0 initially, E = 0 always since energy is conserved.

y

y=0

See text: 8-3


Physics II, Pg 66


E = 1/2mv2 + mgy. So at y = -h, E = 1/2mv2 - mgh = 0. 1/2mv2 = mgh 1/2mv2 will be maximum when mgh is minimum. 1/2mv2 will be maximum at the bottom of the swing !

y

y=0y=-h

h

See text: 8-3


Physics II, Pg 67


1/2mv2 will be maximum at the bottom of the swing ! So at y = -h1 1/2mv2 = mgh1 v2 = 2gh1

v

h1

y

y=0

y=-h1

v gh 2 1

See text: 8-3


Physics II, Pg 68


Since 1/2mv2 - mgh = 0 it is clear that the maximum height on the other side will be at y = 0 and v = 0.

The ball returns to it’s original height.

y

y=0

See text: 8-3


Physics II, Pg 69


The ball will oscillate back and forth. The limits on it’s height and speed are a consequence of the sharing of energy between K and U.

E = 1/2mv2 + mgy = K + U = 0.

y

See text: 8-3

See example A Pendulum

Physics II, Pg 70

Example: Airtrack & GliderExample: Airtrack & Glider

A glider of mass M is initially at rest on a horizontal frictionless track. A mass m is attached to it with a massless string hung over a massless pulley as shown. What is the speed v of M after m has fallen a distance d ?

d

M

m

v

v


Physics II, Pg 71

Example: Airtrack & GliderExample: Airtrack & Glider

Energy is conserved Energy is conserved since all forces are conservative. Choose initial configuration to have U=0.

K = -U

d

M

m

v

1

22m M v mgd lets check this


Physics II, Pg 72

Problem: Hotwheel.Problem: Hotwheel.

A toy car slides on the frictionless track shown below. It starts at rest, drops a distance d, moves horizontally at speed v1, rises a distance h, and ends up moving horizontally with speed v2.

Find v1 and v2.

hd

v1

v2

See text: 8-4

Physics II, Pg 73

Problem: Hotwheel...Problem: Hotwheel...

Energy is conserved, so E = 0 K = - U Moving down a distance d, U = -mgd, K = 1/2mv1

2

Solving for the speed:

hd

v1

v gd1 2

See text: 8-4

Physics II, Pg 74

Problem: Hotwheel...Problem: Hotwheel...

At the end, we are a distance d-h below our starting point. U = -mg(d-h), K = 1/2mv2

2

Solving for the speed:

hd

v2

v g d h2 2

d-h

See text: 8-4

Physics II, Pg 75

Non-conservative Forces:Non-conservative Forces:

If the work done does not depend on the path taken, the force involved is said to be conservative.

If the work done does depend on the path taken, the force involved is said to be non-conservative.

An example of a non-conservative force is friction:

Pushing a box across the floor, the amount of work that is done by friction depends on the path taken.Work done is proportional to the length of the path !

See text: 8-6

Physics II, Pg 76

Non-conservative Forces: FrictionNon-conservative Forces: Friction

Suppose you are pushing a box across a flat floor. The mass of the box is m and the kinetic coefficient of friction is . The work done in pushing it a distance D is given by:

Wf = FFf ..DD = -mgD.

D

Ff = -mg

See text: 8-6

See example 8-9

Physics II, Pg 77

Non-conservative Forces: FrictionNon-conservative Forces: Friction

Since the force is constant in magnitude, and opposite in direction to the displacement, the work done in pushing the box through an arbitrary path of length L is just Wf = -mgL.

Clearly, the work done depends on the path taken.

Wpath 2 > Wpath 1.

A

B

path 1

path 2

See text: 8-6

Physics II, Pg 78

Generalized Work Energy Theorem:Generalized Work Energy Theorem:

Suppose FNET = FC + FNC (sum of conservative and non-conservative forces).

The total work done is: WTOT = WC + WNC

The Work Kinetic-Energy theorem says that: WTOT = K.

WTOT = WC + WNC = K

But WC = -U

So WNC = K + U = E

Physics II, Pg 79

Generalized Work Energy Theorem:Generalized Work Energy Theorem:

The change in total energy of a system is equal to the work done on it by non-conservative forces. E of system not conserved !

If all the forces are conservative, we know that energy is conserved: K + U = E = 0 which says that WNC = 0,which makes sense.

If some non-conservative force (like friction) does work,energy will not be conserved by an amount equal to this work, which also makes sense.

WNC = K + U = E

Physics II, Pg 80

Problem: Block Sliding with FrictionProblem: Block Sliding with Friction

A block slides down a frictionless ramp. Suppose the horizontal (bottom) portion of the track is rough, such that the kinetic coefficient of friction between the block and the track is . How far, x, does the block go along the bottom portion of

the track before stopping ?

x

d

See text: 8-6

Physics II, Pg 81

Problem: Block Sliding with Friction...Problem: Block Sliding with Friction...

Using WNC = K + U As before, U = -mgd WNC = work done by friction = -mgx. K = 0 since the block starts out and ends up at rest. WNC = U -mgx = -mgd

x = d /

x

d

See text: 8-6

Physics II, Pg 82


Work done by variable force in 3-D. (Text: 7-4) Newtons gravitational force. (Text: 8-1, 9-2, 9-4)

Conservative Forces & Potential energy. (Text: 8-1 to 8-5) Conservation of “Total Mechanical Energy”

Examples: Pendulum, Airtrack, Hotwheel. Nonconservative force (Text: 8-1 to 8-6)

friction General work-energy theorem Example problem.

Look at Textbook problemsLook at Textbook problems Chapter 8: 1,2,5,9,10,11,14,15,17 Chapter 8: 1,2,5,9,10,11,14,15,17

Physics II, Pg 83

Physics IIPhysics II


Problems using work-energy theorem. Spring shot. Escape velocity. Loop the loop. Vertical springs.

Definition of Power, with example

Physics II, Pg 84

Problem: Spring ShotProblem: Spring Shot

A sling shot is made from a pair of springs each having spring constant k. The initial length of each spring is x0. A puck of mass m is placed at the point connecting the two springs and pulled back so that the length of each spring is x1. The puck is released. What is it’s speed v after leaving the springs? (The relaxed length of each spring is xR).

v

x0

x1

m m m

xR

See text: 7-3

Physics II, Pg 85


Only conservative forces are at work, so energy is conserved. EI = EF K = -Us

U k x x x x k x x x xs R R R R 21

2 02

12

02

12

x0

x1

m m

See text: 7-3

Physics II, Pg 86



K mv1

22

vm mat rest

See text: 7-3

Physics II, Pg 87



vm m

1

22

02

12mv k x x x xR R

See text: 7-3

Physics II, Pg 88

Problem: How High?Problem: How High?

A projectile of mass m is launched straight up from the surface of the earth with initial speed v0. What is the maximum distance from the center of the earth RMAX it reaches before falling back down.

RMAX

RE v0

m

M


Physics II, Pg 89

Problem: How High...Problem: How High...

No non-conservative forces:WNC = 0 K = -U

RMAX

v0

m

hMAX

U GMmR RE MAX

1 1

And we know:

K mv1

2 02

RE

M

See text: 9-4

Physics II, Pg 90

1

2

1 1

21 1

2 1

2 1

12

02

02

2

02

mv GMmR R

v GMR R

GM

RR

R

R

gRR

R

R

R

v

gR

E MAX

E MAX

EE

E

MAX

EE

MAX

E

MAX E

RMAX

RE v0

m

hMAX

M

RR

v

gR

MAXE

E

1

202

Problem: How High...Problem: How High...

See text: 9-4

Physics II, Pg 91

Escape VelocityEscape Velocity

If we want the projectile to escape to infinity we need to make the denominator in the above equation zero:

RR

v

gR

MAXE

E

1

202

12

002

v

gRE

v

gRE

02

21 v gRE0 2

We call this value of v0 the escape velocity vesc

See text: 9-4

Physics II, Pg 92

Escape VelocityEscape Velocity

Remembering that we find the escape velocity from

a planet of mass Mp and radius Rp to be:

(where G = 6.67 x 10-11 m3 kg-1 s-2). v

GM

Rescp

p

2

Moon

Earth

Sun

Jupiter

Rp(m)

gGM

RE

2

Mp(kg) gp(m/s2) vesc(m/s)

6.378(10)6 5.977(10)24

1.738(10)6 7.352(10)22

7.150(10)7 1.900(10)27

6.960(10)8 1.989(10)29

9.80

1.62

24.8

27.4

11.2(10)3

2.38(10)3

59.5(10)3

195.(10)3

Physics II, Pg 93

Problem: Space SpringProblem: Space Spring

A low budget space program decides to launch a 10,000 kg space-ship into space using a big spring. If the space ship is to reach a height RE above the surface of the earth, what distance d must the launching spring be compressed if it has a spring constant of 108 N/m.

d

k


Physics II, Pg 94

Problem: Space Spring...Problem: Space Spring...

Since gravity is a conservative force, energy is conserved. Since K = 0 both initially and at the maximum height (v=0) we know:

Ubefore = Uafter

{ US + UG }before = { UG }after

1

2 22kd

GMm

R

GMm

RE E

1

2 2

2

2 22kd

GMm

R

GMm

R

GMm

RE E E

dGMm

kRE


Physics II, Pg 95

Problem: Space SpringProblem: Space Spring

For the numbers given, d = 35.3 m

But don’t get too happy...F = kd = maa = kd/ma = 1.8 x 106 m/s2.a = 180000 gastronaut unhappy

dGMm

kRE

d

k

a

So we find


Physics II, Pg 96

Problem: Loop the loopProblem: Loop the loop

A mass m starts at rest on a frictionless track a distance H above the floor. It slides down to the level of the floor where it encounters a loop of radius R. What is H if the mass just barely makes it around the loop without losing contact with the track.

HR

Physics II, Pg 97


Draw a FBD of the mass at the top of the loop:

FFTOT = -(mg+N) j j maa = -mv2/R j j If it “just” makes it, N = 0.

mg = mv2/R

HR

v

mg

N

v

i i

j j

v Rg

Physics II, Pg 98


Now notice that energy is conserved. K = -U. U = -mg(h) = -mg(H-2R), K = 1/2 mv2 = 1/2 mRg

mg(H-2R) = 1/2 mRg

HR

h = H - 2R

v

H R5

2

Physics II, Pg 99

Vertical SpringsVertical Springs

A spring is hung vertically, it’s relaxed position at y=0 (a). When a mass m is hung from it’s end, the new equilibrium position is yE (b).

Free Body Diagram of mass in case (b) lets us relate k and yE:

-kyE - mg = 0 (yE < 0)

mg = -kyE

y = 0

y = yE

j j

k

m

(a) (b)

(ok since yE is a negative number)

See text: 7-3

Recall example 7-7, Bungee Jumping

Physics II, Pg 100


The potential energy of the spring-mass system is:

y = 0

y = yE

j j

k

m

(a) (b)

U ky mgy C 1

22

U ky ky y CE 1

22

but mg = -kyE

choose C to make U=0 at y = yE:

01

22 2 ky ky CE E C kyE

1

22

Recall example 7-7, Bungee Jumping

See text: 7-3, 8-2

Physics II, Pg 101


So:

y = 0

y = yE

j j

k

m

(a) (b)

U ky ky y k

k y y y y

E E

E E

1

2

1

21

22

2 2

2 2

which can be written:

U k y yE 1

22

See text: 7-3, 8-2

Physics II, Pg 102


So if we define a new y’ co-ordinate system such that y’ = 0 is at the equilibrium position, ( y’ = y - yE ) then we get the simple result:

y’ = 0

j j

k

m

(a) (b) U k y yE

1

22

U ky1

22'

See text: 7-3, 8-2

Physics II, Pg 103


If we choose y = 0 to be at the equilibrium position of the mass hanging on the spring, we can define the potential in the simple form.

Notice that g does not appear in this expression !!By choosing our coordinates and constants cleverly, we can hide

the effects of gravity.

y = 0

j j

k

m

(a) (b)

U ky1

22

See text: 7-3, 8-2

Physics II, Pg 104

U of Spring

-60

-40

-20

0

20

40

60

80

100

120

140

160

-10 -8 -6 -4 -2 0 2 4 6 8 10

y

U US = 1/2ky2

See text: 8-2

See Figure 8-9

Physics II, Pg 105

U of Gravity

-60

-40

-20

0

20

40

60

80

100

120

140

160

-10 -8 -6 -4 -2 0 2 4 6 8 10

y

U

UG = mgy

See text: 8-2

See Figure 8-7

Physics II, Pg 106

U of Spring + Gravity

-60

-40

-20

0

20

40

60

80

100

120

140

160

-10 -8 -6 -4 -2 0 2 4 6 8 10

y

U

UG = mgy

US = 1/2ky2

UTOT =UG + US

shift due to mgy term

See text: 8-2

Physics II, Pg 107

U of Spring + Gravity

-60

-40

-20

0

20

40

60

80

100

120

140

160

-10 -8 -6 -4 -2 0 2 4 6 8 10

y

U US = 1/2ky2

UTOT =UG + US + C

shift due to mgy term

See text: 8-2

Physics II, Pg 108

Vertical Springs:Vertical Springs:Example ProblemExample Problem

If we displace the mass a distance d from equilibrium and let it go, it will oscillate up & down. Relate the maximum speed of the mass vM to d and the spring constant k.

Since all forces are conservative,E = K + U is constant.We know:

y = 0

j j

k

m

U ky1

22

y = d

y = -d

vMK mv1

22

See text: 7-3, 8-2

Physics II, Pg 109

Vertical Springs:Vertical Springs:Example ProblemExample Problem

At the initial stretched positionand K = 0 (since v=0).

Since E is conserved,will always be true !

Energy is shared between the K and U terms. At y = d or -d the energy is all potentialAt y = 0, the energy is all kinetic.

y = 0

j j

k

mE kd ky mv

1

2

1

2

1

22 2 2

y = d

y = -d

vM

U kd1

22

E kd1

22

1

2

1

22 2kd mv v d

k

m

See text: 7-3, 8-2

Physics II, Pg 110

PowerPower We have seen that W = FF.dSS

This does not depend on time !

Power is the “rate of doing work”:

If the force does not depend on

time: dW/dt = FF.dSS/dt = FF.v v P = FF.vv

Units of power: J/sec = Nm/sec = Watts

PdW

dt

FFSS

vv

See text: 7-6

Physics II, Pg 111

PowerPower A 2000kg trolley is pulled up

a 30 degree hill at 20 mi/hrby a winch at the top of thehill. How much power must thewinch be producing ?

The power is P = FF.v v = TT.v v Since the trolley is not accelerating, the net force on it must be zero. In the x direction:

T - mg sin() = 0T = mg sin()

vv

mg

TT

winch

xy

See text: 7-6

Physics II, Pg 112

PowerPower P = TT.vv = Tv

since TT is parallel to vv

So P = mgv sin()

v = 20 mi/hr = 8.93 m/sg = 9.8 m/s2

m = 2000 kgsin() = sin(30o) = 0.5

and P = (2000 kg)(9.8 m/s2)(8.93 m/s)(0.5) = 88,000. W

vv

mg

TT

winch

xy

See text: 7-6

Physics II, Pg 113


Problems using work-energy theorem. Spring shot. Escape velocity. Loop the loop.Vertical springs.

Definition of Power, with example

Look at textbook problems Look at textbook problems

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