PHYS 152

Fall 2005

Thursday, Sep 22

Chapter 6 - Work, Kinetic Energy & Power

Exam #1• Thursday, September 29, 7:00 – 8:00 PM.

• Material: Chapters 2-6.3• Use black lead #2 pencil and calculator.• Formula sheet provided. You may bring 1 extra

sheet of handwritten notes (both sides) but no other materials

• Sample exams on Web (link on homepage)

Room assignments - to be announced.

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Supplemental Instruction

Exam Review Session

Tuesday Sept. 27th

7-9pm

RHPH 172

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Reading Quiz: Sections 6.2 & 6.3

During a short time interval a particle moves along a straight line a distance

sr=2i$+ 2 j$

During that time a constant force acted on the particle:

Fur=4 j$

The work done on the particle was

a. 0

b. 4

c. 8

d. 16

e. .8 2

Pg 5

Two marbles, one twice as heavy as the the other, are dropped to the ground from the roof a building. Just before hitting the ground, the heavier marble has

(a) as much kinetic energy as the lighter one

(b) twice as much kinetic energy as the lighter one

(c) half as much kinetic energy as the lighter one

(d) four times as much kinetic energy as the lighter

one

(e) impossible to determine

Pg 6

Two marbles, one twice as heavy as the the other, are dropped to the ground from the roof a building. Just before hitting the ground, the heavier marble has

(a) as much kinetic energy as the lighter one(b) twice as much kinetic energy as the lighter one(c) half as much kinetic energy as the lighter one(d) four times as much kinetic energy as the lighter one(e) impossible to determine

€

W = ΔK = Fd

ΔK1 = Fd = m1gd

ΔK2 = Fd = 2m1gd

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W = FF rr

• If = 90o no work is done.

Review: Constant Force...

vvNN

TT

vv

No work done by NN

No work done by TT

Pg 8

Work-Kinetic Energy Theorem

€

W = ΔK =12

mv22 −

12

mv12

Work done by the net external (constant) force equals the change in kinetic energy

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

Pg 9

• First calculate the work done by gravity:

Wg = mgg rr = -mg rr

• Now find the work done bythe hand:

WHAND = FFHAND rr = FHAND rr

Work done by Lifting Example: Lifting a book from the floor to a shelf

mgg

rr FFHAND

vv = const

aa = 0

floor

shelf

Pg 10

Work done by Lifting

Work/Kinetic Energy Theorem: W = K

€

K = K f − Ki = WNET

When lifting a book from the floor to a shelf, the object is stationary before and after the lift:

€

Ki = K f = 0, ΔK = 0, WNET = 0

WNET= 0

WNET = WHAND + Wg

= FHAND rr - mg rr

= (FHAND - mg) rr

WHAND = - Wg

mgg

rr FFHAND

vv = const

aa = 0

floor

shelf

Pg 11

Lifting vs. Lowering

WHAND = - Wg

mgg

rr FFHAND

vv = const

aa = 0

floor

shelf

mgg

rr FFHAND

vv = const

aa = 0

floor

shelfLifting Lowering

Wg = -mg rr

WHAND = FHAND rr

WHAND = - Wg

Wg = mg rr

WHAND = -FHAND rr

€

WNET = 0

Pg 12

rr11rr22

rr33

rrnn

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

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

Work done by gravity...

m

mgg

h j j

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

Wg = mg h

r

= F r

= F y

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Work done by Variable Force: (1D)* When the force was constant, we wrote W = F x

area under F vs. x plot:F

x

Wg

x

€

W = F(x)dxx1

x2

∫

F(x)

x1 x2 dx

* For variable force, we find the area by integrating:

dW = F(x) dx.

Pg 14

€

W =x1

x2

∫ dx

Work/Kinetic Energy Theorem for a Variable Force

dtmma ==FF

dx

€

=mdt

x1

x2

∫ dv

€

=mv1

v2

∫v dv

€

=m12

( − ) =12

m −12

m = ΔKEv22 v1

2 v22 v1

2

dv

dxdxdv dv dv

dxv (chain rule)

€

=mv1

v2

∫ dxdv

dxv

dt=

dt=

Pg 15

Power is the rate at which work is done by a forcePower is the rate at which work is done by a force

PPAVGAVG = W/ = W/t Average Powert Average Power

P = dW/dt Instantaneous PowerP = dW/dt Instantaneous Power

The unit of power is a Joule/second (J/s) which we define as a The unit of power is a Joule/second (J/s) which we define as a Watt (W)Watt (W)

1 W = 1 J/s1 W = 1 J/s

P =

dWdt

=ddt

rF ⋅d

rx( ) =

ddt

(Fur

gvrdt) =

rF ⋅

rv

Power

Pg 16

The force on a particle of mass m is given by

Fur=−10xi$

Choose the correct statement:

a. The work done will be the same going from x=1 to x=2 as it is for going from x=0 to x=1.

b. The work done will be the same going from x=1 to x=2 as it is for going from x=-1 to x=-2.

c. The average power is the same as the instantaneous power.

d. None of the above are correct.

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Work done by a Spring

*For a person to hold a spring stretched out or compressed by x from its unstretched length, it requires a force

where k =spring constant measures the stiffness of the spring.

Spring unstretchedx=0

Fs Fp Person pulling

FsFp

Person pushing

kxFp =

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Work done by a Spring

The spring exerts a force (restoring force) in the opposite direction:

where k =spring constant measures the stiffness of the spring.

Spring unstretchedx=0

Fs Fp Person pulling

FsFp

Person pushing

kxFs −= Hooke’s law

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1-D Variable Force Example: Spring

*For a spring Fx = -kx. ( Hooke’s Law)

k =spring constant

F(x) x2

x

x1

-kxrelaxed position

F = - k x1

F = - k x2

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1-D Variable Force Example: 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.

F(x) x2

x

x1

-kxrelaxed position

F = - k x1

F = - k x2

Ws

Pg 21

1-D Variable Force Example: 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.

F(x) x2

x

x1

-kx

Ws

€

Ws = F(x)dxx1

x2

∫

= (−kx)dxx1

x2

∫

= −12

kx2

x1

x2

Ws = −12

k x22 − x1

2( )

Pg 22

Work - EnergyA 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.

xv1

m1

m1

Pg 23

Work - Energy

x1v1

m1

m1

Use the fact that WNET = K.

so kx2 = mv2k

mvx 1

11=

In this caseWNET = WSPRING = -1/2 kx2 and K = -1/2 mv2

In the case of x1

Pg 24

Work - Energy

x2v2

m2

m2

If the initial speed of the box were doubled and its mass were halved, how far x2 would the spring compress ?

(a)(a)12

xx =12

x2x =(b)(b) 12x2x =(c)(c)

Pg 25

Work - Energy

x2v2

m2

m2

If the initial speed of the box were doubled and its mass were halved, how far x2 would the spring compress ?

k

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

k

2mv

k

2mv2x 1

11

12==

12x2x =

Pg 26

ExampleA person pulls on a spring. It requires a force of 75N to stretch it by 3 cm. How much work does the person do? If the person compresses the spring by 3 cm how much work

does the person do? Calculate the spring constant:

The work is

The work to compress the spring is the same since W is proportional to x 2.

€

k =Fx

=75N

0.03m= 2.5 ×103 N / m

€

W =12

kxmax2 =

12

2.5 ×103 N / m( ) 0.03m( )2 = 1.1J

Pg 27

Example: Compressed SpringA horizontal spring has k=360N/m. (a) How much work is required to compress a spring from x=0 to x=-11 cm? (b) If a 1.85 kg block is placed against the spring and the spring is released what will be the speed of the block when it separates from the spring at x=0?

mg x=0

Fs=kx

N

x=-11

Pg 28

Example: Compressed SpringThe work done to stretch or compress the spring is:

In returning to its uncompressed length the spring will do work W=2.18J on the block.

According to the work-energy principle the block acquires kinetic energy:

( )( ) J18.2m11.0m/N3602

1kx

2

1W 22 ===

€

Wnet = K f − Ki =12

mv2 − 0 ΔK =12

mv2

€

v =2(ΔK )

m=

2 2.18J( )1.85Kg

= 1.54m / s