Ari (Energy) Kuliah Ke-2

7/27/2019 Ari (Energy) Kuliah Ke-2

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PHYSICS DEPARTMENT

19 September 2013


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Work and Ener


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Work.

order for work to take place, a force must be exertedthrou h a distance. The amount of work done de endson two things: the amount of force exerted and the

distance over which the force is applied. There aretwo factors to keep in mind when deciding when workis being done: something has to move and the motion

.can be calculated by using the following formula:Work=force x distance


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negative

Man does positive worklifting box

Man does negative work

lowering box

Gravity does positive work

when box lowers

Gravity does negative

work when box is raised


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Work done by a constant Force

W = F s = |F| |s| cos = Fs sF

|F| : magnitude of force|s| = s : magnitude of displacement

s

Fs = magnitude of force in

direction of displacement :

Fs = |F| cos

: ang e e ween sp acemen an orcevectors

Kinetic energy : Ekin= 1/2 m v2

Ekin = Wnet Work-Kinetic Energy Theorem:


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Work Done by Gravity

Wg = (mg)(S)cosh0

= cosWg = mg(h/cos)cosW = mgh

h

mg

S

hwith h= h0-hf

Work done by gravity is independent of path

0 f

=> The gravitational force is a conservative force.


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Concept Question

Imagine that you are comparing three different ways of having a ball move

down through the same height. In which case does the ball reach the

bottom with the highest speed?

1. Dropping2. Slide on ramp (no friction)

3. Swin in down

4. All the same 1 2 3correct

In all three experiments, the balls fall from thesame hei ht and therefore the same amount of their

gravitational potential energy is converted tokinetic energy. If their kinetic energies are all the

, ,all have the same speed at the end.


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Work-Kinetic Energy Theorem

Work done by the net external (constant) force

equals the change in kinetic energy

W = K =2

mv2

2 2

mv1

2

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


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Work done by Lifting Example: Lifting a book

from the floor to a shelf

First calculate the work done b ravit :

W = m rr= -m rrshelf

rr FFHAND

vv = constant=

the hand:mgg

WHAND = FFHAND rr= FHAND rrfloor


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Work done by Lifting shelf

Work/Kinetic Energy Theorem: W = Krr HAND

vv = const

aa = 0

K = Kf Ki = WNET mgg

When lifting a book from the floor to a shelf, the object is

Ki = Kf = 0, K = 0, WNET = 0

NET=WNET = WHAND + Wg

= -WHAND = - Wg

= (FHAND

- mg) rr


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Lifting vs. Lowering

shelf shelfLifting Lowering

rr HANDvv = const

aa = 0

rr HANDvv = const

aa = 0

mgg

mggoor

Wg = -mg rr Wg = mg rr

WHAND = FHAND rr WHAND = -FHAND rr

WHAND

= - Wg WHAND = - WgWNET = 0


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Work done by gravity

=

= FF rr1+ FF rr2 + . . . +FF rrn= FF rr + rr + . . .+ rr m

. . . n

rr11rr22

mgg

h = F r= F y

W = m h

rrnn


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

area underF vs. x plot:F

x

Wg

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

xF(x)

dW = F(x) dx.

W = F(x)dxx

x1 x2dx


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Work (Kinetic Energy) Theorem for a

x2

x1 dt

mma ==F

dxx2

dvdxdv dv dv

v (chain rule)= =

dtx1

v2

v2

= mv1

dxxv

= mv1

v dv

1 1 1= m

2 =

2m

2m =v2 v1 v2 v1


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Power

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/tt Average Power Average Power

P =P = dWdW//dtdt Instantaneous Power Instantaneous Power

The unit of ower is a Joule/second J/s whichThe unit of ower is a Joule/second J/s which

we define as a Watt (W)we define as a Watt (W)


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Work done by a Spring

x=0Spring unstretched

Fs Fp Person

pullingPerson

* For a person to hold a spring stretched out or compressed by

sp

,

kxFp =w ere =spr ng cons an measures e s ness o espring.


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Work done by a Spring

x=0Spring unstretched

Fs Fp Person

pullingPerson

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

sp

kxFs Hookes law

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


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Variable Force Example: Spring

*For a spring Fx = -kx. ( Hookes Law)

k =s rin constant

F x

x

-kxrelaxed position

F = - k x1

- 2


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* The work done by the spring Ws during a displacementfrom x1 to x2 is the area under the F(x) vs x plot between

x1 an x2.

F x

x

-kxrelaxed position

s

F = - k x1

- 2


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*The work done by the springWs during a displacement from

Ws = F(x )dxx 2

x1 o x2 s e area un er e

F(x) vs x plot betweenx and x . = kx dx

x2

x1

1 x 2

F(x) x2x1

= 2

xx1

x

-

WsWs =

1k x 2

2 x12


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Work - Energy

fixed spring, compressing it a distance x1 from its relaxedposition while momentarily coming to rest.

xv

m1

m1


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Work - Energy

Use the fact that WNET = K.

In this case

WNET = WSPRING = -1/2 kx

2 and K = -1/2 mv2

so kx2

= mv2

k

mvx 1

11=

In the case of x1

x1v

m1

m1


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Work - Energy

were halved, how far x2 would the spring compress ?

(a)(a) 12 xx = 12 x2x =(b)(b) 12 x2x =(c)(c)

x2v

m2

m2


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Work - Energy

were halved, how far x2 would the spring compress ?

kvx =

o v2 = v1 an m2 = m1

2m2m

kv

kvx

112==

x2v

12=

m2

m2


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Example

.stretch it by 3 cm. How much work does the person do? Ifthe person compresses the spring by 3 cm how much workdoes the erson do?

Calculate the spring constant:

k=x

=

0.03m

= 2.5 10 N/ m

The work is

1 1 2

= 2 xmax = 2 . m . m = .

proportional to x2.


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Example: Compressed Spring

.work is required to compress a spring from x=0 tox=-11 cm? (b) If a 1.85 kg block is placed against

speed of the block when it separates from the springat x=0?

Fs=kx

mg x=0x=-11


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Example: Compressed Spring

( )( )11 22 =In returnin to its uncom ressed len th the s rin will do

( )( )22

work W=2.18J on the block.

-kinetic energy:

W = K K =1

mv2 0 K =

1mv

2

v =m

=.

1.85Kg

= 1.54m / s


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Ari (Energy) Kuliah Ke-2