Physics 151: Lecture 12, Pg 1 Physics 151: Lecture 12 l Topics : ç( Text Ch. 7.1-7.4) »Work & Energy. »Work of a constant force. »Work of a non-constant

Physics 151: Lecture 12, Pg 1

Physics 151: Lecture 12Physics 151: Lecture 12

Topics :(Text Ch. 7.1-7.4)

» Work & Energy.

» Work of a constant force.

» Work of a non-constant force.» Work - Energy Theorem.


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 (thermal energy).


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 an otherwise isolated system will change it’s “energyenergy”...

See text: 7-1

Animation


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 = F F . rr = F rr cos = Fr rr

FF

rr

displace

ment

Fr

“Dot Product”

See text: 7-1


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

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

FF

rr

F cos


Review: Scalar Product ( or Dot Product)Review: Scalar Product ( or Dot Product)

Definition:

aa.bb = ab cos

= a[b cos ] = aba

= b[a cos ] = bab

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

The dot product of perpendicular vectors is 0 !!

See text: 7.2

aa

ab bb

aa

bb

ba


Review: Examples of dot productsReview: 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

x

y

z

ii

jj

kk


Review: Properties of dot productsReview: 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

Pythagorian Theorem !!

aa

ax

ay

ii

j j

See text: 7.2


Review: Properties of dot productsReview: 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


Lecture 12, Lecture 12, ACT 1ACT 1WorkWork

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

Is the work done by F positive or negative?


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

WF = (10 N)x(5m) = 50 N-m.

See text: 7-1

See example 7-1: Pushing a trunk.


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


Work and Varying ForcesWork and Varying Forces

Consider a varying force,

W = Fxx

As x 0, x dx

Text : 7.3

Fx

xx

Area = Fxx

dxFW

dxFdW

x

x


SpringsSprings

A very common problem with a variable force is a spring.

In this spring, the force gets greater as the spring is further compressed.

Hook’s Law,

FS = - k x

x is the amount the spring is stretched or compressed from it resting position.

Text : 7.3

Fx

Animation


Lecture 12,Lecture 12, ACT 2ACT 2Hook’s LawHook’s Law

Remember Hook’s Law,

Fx = -k x

What are the units for the constant k ?

A) B) C) D)2

2

smkg

2smkg

2skg

2

2

smkg


Lecture 12,Lecture 12, ACT 3ACT 3Hook’s LawHook’s Law

0.2 kg

9 cm

8 cm

What is k for this spring ??

A) 50 N/m B) 100 N/m C) 200 N/m D) 400 N/m


What is the Work done by the Spring...What is the Work done by the Spring...

x2 x1

F(x)

x

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.

2

1

2

2

2

1

2

2s

2

21

21

W

21

)(

)(

2

1

2

1

2

1

xxkWU

xxk

kx

dxkx

dxxFW

s

x

x

x

x

x

xs

Ws

kx1

kx2

-kx

Ws

Ws = - 1/2 [ ( kx2) (x2) - (kx1) (x1) ]


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


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

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

W = (F).d = ma.dFor constant a, a = (v-v0)/t

also, d = vavt = 1/2 (v+v0)t

xx

FFv1 v2

aa

ii

m


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

Altogether,

W = (F).d = ma.d = m [(v-v0)/t) .(1/2 (v+v0)t]

W = 1/2 m ( v2 - v02 ) = (1/2 m v2 ) - (1/2 m v0

2 )

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


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

NetNet WorkWork done on object = changechange in kinetic energykinetic energy of object

KWnet

See text: 7-4

12 KK

21

22 mv

21

mv21

Is this applicable also for a variable force ? YES


Lecture 12, Lecture 12, ACT 4ACT 4Kinetic EnergyKinetic Energy

To practice your pitching you use two baseballs. The first time you throw a slow curve and clock the speed at 50 mph (~25 m/s). The second time you go with high heat and the radar gun clocks the pitch at 100 mph. What is the ratio of the kinetic energy of the fast ball versus the curve ball ?

(a)(a) 1/4 (b)(b) 1/2 (c)(c) 1 (d) 2 (e) 4


xx

vo

m

to

F

ExampleExampleWork Kinetic-Energy TheoremWork Kinetic-Energy Theorem

x = ( m vo2 / k ) 1/2

How much will the spring compress to bring the object to a stop if the object is moving initially at a constant velocity (vo) on frictionless surface as shown below ?

spring compressed

spring at an equilibrium position

V=0

t m


Act 4Act 4

Two clowns are launched from the same spring-loaded circus cannon with the spring compressed the same distance each time. Clown A has a 40-kg mass; clown B a 60-kg mass. The relation between their kinetic energies at the instant of launch is :

a) KA = 4 KB

b) KA = 2 KB

c) KA = KB

d) KB = 2 KA

e) KB = 4 KA


Act 4.bAct 4.b

Two clowns are launched from the same spring-loaded circus cannon with the spring compressed the same distance each time. Clown A has a 40-kg mass; clown B a 60-kg mass. The relation between their speeds at the instant of launch is:

a) vA = 3/2 vB

b) vA = (3/2 )1/2 vB

c) vA = vB

d) vB = (3/2 )1/2 vA

e) vB = 3/2 vA


Recap of today’s lectureRecap of today’s lecture Work & Energy. (Text: 7.1-4)

Discussion.Definition.

Work of a constant and non-constant forces. Work - Energy Theorem.

Documents

Physics 151: Lecture 12, Pg 1 Physics 151: Lecture 12 l Topics : ç( Text Ch. 7.1-7.4) »Work & Energy. »Work of a constant force. »Work of a non-constant