Mats a Selen-Phys111-Lec9

Physics 111: Lecture 9, Pg 1

Physics 111: Lecture 9Physics 111: Lecture 9

Todays AgendaTodays Agenda

l Work & Energy Discussion Definition

l Dot Productl Work of a constant force

Work/kinetic energy theoreml Work of multiple constant forcesl Comments


Work & EnergyWork & Energy

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

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

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

solve problems


Forms of EnergyForms of Energy

ll 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 (thermal energy).


Mass = Energy Mass = Energy (but not in Physics 111)(but not in Physics 111)

l Particle Physics:

+ 5,000,000,000 V

e-

- 5,000,000,000 V

e+(a)

(b)

(c)

E = 1010 eV

M E = MC2

( poof ! )


Energy ConservationEnergy Conservationl Energy cannot be destroyed or created.

Just changed from one form to another.

l 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.

l Doing workwork on an isolated system will change its energyenergy...

ReturningCan

Wilberforce


Definition of Work:Definition of Work:

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

Work, W, of a constant force FFacting through a displacement Drris:W = FFii Drr = F Drr cos q = Fr Drr

q

FF

DrrFr

Dot Product


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

l Only the component of F along the displacement is doing work. Example: Train on a track.

FF

D rrqF cos q

Hairdryer


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

aa.bb = ab cos q= a[b cos q] = aba

= b[a cos q] = bab

Some properties:aaii bb = bbii aaq(aaii bb) = (qbb)ii a a = bbii (qaa) (q is a scalar)aaii (b b + cc) = (aaii bb) + (aaii cc) (cc is a vector)

The dot product of perpendicular vectors is 0 !!

q

aa

ab bb

q

aa

bb

ba


Aside: Examples of dot productsAside: Examples of dot products

Suppose Then

aa = 1 i i + 2 j j + 3 kkbb = 4 i i - 5 j j + 6 kk

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 = 1i i . jj = j j . k k = k k . i i = 0

x

y

zii

jj

kk


Aside: Properties of dot productsAside: Properties of dot products

l 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

l Components:aa = ax i i + ay j j + az kk = (ax , ay , az) = (aa . ii, aa . jj, aa . kk)

l Derivatives:

Apply to velocity

So if v is constant (like for UCM):

ddt

ddt

ddt

( )a b a b a b = +

ddt

v ddt

ddt

ddt

2 2= = + = ( )v v v v v v 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 D rr is:W = FFii D rr FF

D rr

Skateboard


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

l A box is pulled up a rough (m > 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

l Draw FBD of box:N

f

mg

T l Consider direction of

motion of the box

v

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

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

Dxx

FF

Work done byby F F onon box :WF = FFii Dxx = F Dx (since FF is parallel to Dxx)WF = (10 N) x (5 m) = 50 Joules (J)Joules (J)


Units:Units:

N-m (Joule) Dyne-cm (erg)= 10-7 J

BTU = 1054 Jcalorie = 4.184 Jfoot-lb = 1.356 JeV = 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:

l A force FF = 10 N pushes a box across a frictionlessfloor for a distance Dx x = 5 m. The speed of the box is v1before the push and v2 after the push.

Dxx

FFv1 v2

ii

m


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

l Since the force FF is constant, acceleration aa will be constant. We have shown that for constant a: v22 - v12 = 2a(x2-x1) = 2aDx. multiply by 1/2m: 1/2mv22 - 1/2mv12 = maDx But F = ma 1/2mv22 - 1/2mv12 = FDx

Dxx

FFv1 v2

aa

ii

m


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

l So we find that 1/2mv22 - 1/2mv12 = FDx = WF

l Define Kinetic Energy K: K = 1/2mv2

K2 - K1 = WF WF = DK (Work/kinetic energy theorem)(Work/kinetic energy theorem)

Dxx

FFaa

ii

m

v2v1


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

{NetNet WorkWork done on object}=

{changechange in kinetic energy kinetic energy of object}

KWnet D=

12 KK -=

21

22 mv2

1mv21

-=

l Well prove this for a variable force later.



l 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. m > 0) which slows them down to a stop.Which one will go farther before stopping?

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

m1

m2



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

both N and mg are perpendicular to the blocks motion).

mf

N

mg



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

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

m

D

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



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

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

mm2gD2 = mm1gD1 m2D2 = m1D1

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

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

l Wg = FFii Dr r = mg Drr cos(0) = mgH

Wg = mgH


Wg = mgH = 1/2mv2

Drrmg g

Hj j

v0 = 0

vv gH= 2


What about multiple forces?What about multiple forces?

Suppose FFNET = FF1 + FF2 and thedisplacement is Drr.The work done by each force is:

W1 = FF1ii Dr r W2 = FF2 ii Drr

WTOT = W1 + W2= FF1ii Dr r + FF2ii Drr= (FF1 + FF2 )ii Drr

WTOT = FFTOTii Drr Its the totaltotal force that matters!!

FFNETDrrFF1

FF2


Comments:Comments:

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

Since W = FFii Drr

l No work is done if: FF = 0 or Drr = 0 or q = 90o


Comments...Comments...

W = FFii Drr

l No work done if q = 90o.

No work done by TT.

No work done by N.

TT

v v

vvNN



l An inclined plane is accelerating with constant acceleration a. A box resting on the plane is held in place by static friction. How many forces are doing work on the block?

aa

(a)(a) 1 (b)(b) 2 (c) (c) 33



l First, draw all the forces in the system:

aa

mgmg NN

FFSS


aa

mgmg NN


l Recall that W = FFii rr so only forces that have a component along the direction of the displacement are doing work.

FFSS

l The answer is (b) 2.


Recap of todays lectureRecap of todays lecture

l Work & Energy Discussion Definition

l Dot Productl Work of a constant force

Work/kinetic energy theoreml Properties (units, time independence, etc.)l Work of a multiple forcesl Comments


Physics 111: Lecture 10Physics 111: Lecture 10

Todays AgendaTodays Agenda

l Review of Workl Work done by gravity near the Earths surfacel Examples:

pendulum, inclined plane, free falll Work done by variable force

Springl Problem involving spring & friction


Review: Constant ForceReview: Constant Force

Work, W, of a constant force FFacting through a displacement Drris:W = FFii Dr r = F Dr cos(q) = Fr Drr

q

FF

DrrFr


Review: Sum of Constant ForcesReview: Sum of Constant Forces

Suppose FFNET = FF1 + FF2 and thedisplacement is SS.

The work done by each force is:

W1 = FF1ii Dr r W2 = FF2ii Dr

FFTOTDrrFF1

FF2 WNET = W1 + W2= FF1ii Dr r + FF2ii Drr= (FF1 + FF2 )ii Drr

WNET = FFNET ii Drr


Review: Constant Force...Review: Constant Force...

W = FFii Drr

l No work done if q = 90o.

No work done by TT.

No work done by NN.

vvNN

TT

v v



{NetNet WorkWork done on object}=

{changechange in kinetic energy kinetic energy of object}

WF = DK = 1/2mv22 - 1/2mv12

Dxx

FFv1 v2

m WF = FDx


Work done by gravity:Work done by gravity:

l Wg = FFii Drr = mg Drr cos q= -mg Dy

Wg = -mg Dy

Depends only on Dy !

j j

m

Drrmg g

-Dyq

m


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

l

Depends only on Dy, not on path taken!

m

mg g

Dy j j

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

Dr

= Fii Dr= F Dy

Drr11Drr22

Drr33

Drrnn

= FFii Drr 1+ FFii Drr2 + . . . + FFii Drrn= FFii (Drr11 + Drr 2+ . . .+ Drrnn)

Wg = -mg Dy


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

l 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 mv12 = 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


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??

l First calculate the work done by gravity:

Wg = mggii Drr = -mg Drr

l Now find the work done bythe hand:

WHAND = FFHANDii Drr = FHAND Drrmgg

Drr FFHANDvv = constaa = 0


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

Wg = -mg DrrWHAND = FHAND Drr

WNET = WHAND + Wg= FHAND Drr - mg Drr= (FHAND - mg) Drr

= 0 since K = 0 (v = const)

l So WTOT = 0!!

mgg


Textbook


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

l Work/Kinetic Energy Theorem says: W = DK

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

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

mgg



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

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

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

F

x

Wg

Dx

=2

1

x

xdx)x(FW

F(x)

x1 x2 dx


Work/Kinetic Energy Theorem for a Work/Kinetic Energy Theorem for a Variable ForceVariable Force

=2

1

x

xW dt

mma ==

KEm21m

21)(

21m =-=-=

FF dx

dx=2

1

x

x dtm dv

dxdv

=2

1

v

vm v dv

v22 v12 v22 v12

dv

dxv

dxdxdv dv dv

dxv (chain rule)

=2

1

v

vm

dt = dt =


11--D Variable Force Example: SpringD Variable Force Example: Spring

l For a spring we know that Fx = -kx.

F(x) x2

x

x1

-kxrelaxed position

F = - k x1

F = - k x2


Spring...Spring...

l 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...

( )2122s

x

x

2

x

x

x

xs

xxk21W

kx21

dxkx

dxxFW

2

1

2

1

2

1

--=

-=

-=

=

)(

)(F(x) x2

Ws

x

x1

-kx

l 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



l 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 =



l Again, use the fact that WNET = DK.

x1v1

so kx2 = mv2km

vx 111 =

m1

m1

In this case, WNET = WSPRING = -1/2 kx2and DK = -1/2 mv2

In the case of x1



x2v2

kmvx =

m2

m2

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

k2m

vk

2mv2x 11

112 ==

12 x2x =


Problem: Spring pulls on mass.Problem: Spring pulls on mass.l 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.l First find the net work done on the mass during the motion

from x = d to x = 0 (only due to the spring):


drelaxed position

vr

m

mii

( ) ( ) 22221

22s kd2

1d0k21xxk

21W =--=--=


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


drelaxed position

vr

m

mii

2r

21

22 mv2

1mv21mv

21

K =-=


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


drelaxed position

vr

m

mii

12

2kd = 2rmv21

mkdv r =


Problem: Spring pulls on mass.Problem: Spring pulls on mass.l Now suppose there is a coefficient of friction m between the

block and the floorl 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.rr = - mmg d


drelaxed position

vr

m

mii

f = mmg

Drr


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

Wf = -mmg d


drelaxed position

vr

m

mii

f = mmg

Drr

W kdS =12

2 2rmv2

1K =D

2r

2 mv21mgdkd

21

=-m gd2dmkv 2r -=


Recap of todays lectureRecap of todays lecture

l Reviewl Work done by gravity near the Earths surfacel Examples:

pendulum, inclined plane, free falll Work done by variable force

Springl Problem involving spring & friction

Mats a Selen-Phys111-Lec9

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