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PHYS 152 Fall 2005 Thursday, Sep 22 Chapter 6 - Work, Kinetic Energy & PowerPowerPoint Presentation

PHYS 152 Fall 2005 Thursday, Sep 22 Chapter 6 - Work, Kinetic Energy & Power

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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.

Reading Quiz: Sections 6.2 & 6.3

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

- The work done on the particle was
- 0
- 4
- 8
- 16
- .

During that time a constant force acted on the particle:

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

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

T dropped to the ground from the roof a building. Just before hitting the ground, the heavier marble has

v

v

N

Review: Constant Force...W = Fr

- If = 90o no work is done.

No work done by T

No work done by N

Work-Kinetic Energy Theorem dropped to the ground from the roof a building. Just before hitting the ground, the heavier marble has

Work done by the net external (constant) force

equals the change in kinetic energy

{NetWork done on object}= {change in kinetic energy of object}

shelf dropped to the ground from the roof a building. Just before hitting the ground, the heavier marble has

FHAND

r

v = const

a= 0

mg

floor

Example: Lifting a book from the floor to a shelf

Work done by Lifting- First calculate the work done by gravity:
Wg = mgr= -mg r

- Now find the work done bythe hand:
WHAND = FHANDr= FHANDr

shelf dropped to the ground from the roof a building. Just before hitting the ground, the heavier marble has

FHAND

r

v = const

a= 0

mg

floor

Work done by LiftingWork/Kinetic Energy Theorem: W = K

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

WNET= 0

WNET = WHAND + Wg

= FHAND r- mg r

= (FHAND- mg)r

WHAND = - Wg

shelf dropped to the ground from the roof a building. Just before hitting the ground, the heavier marble has

FHAND

r

v = const

a= 0

shelf

FHAND

r

v = const

a= 0

mg

floor

mg

floor

Lifting vs. LoweringLifting

Lowering

Wg= -mg r

Wg= mg r

WHAND= FHANDr

WHAND= -FHANDr

WHAND = - Wg

WHAND = - Wg

= dropped to the ground from the roof a building. Just before hitting the ground, the heavier marble hasFr1+ Fr2+ . . . +Frn

= F(r1+ r2+ . . .+ rn)

r1

=Fr

= Fy

r2

r

r3

Wg = mgh

rn

Work done by gravity...W NET = W1 + W2 +. . .+ Wn

m

mg

h

j

F(x) dropped to the ground from the roof a building. Just before hitting the ground, the heavier marble has

x1

x2

dx

Work done by Variable Force: (1D)F

* When the force was constant, we wrote W = Fx

area under F vs. x plot:

Wg

x

x

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

dW = F(x) dx.

dx dropped to the ground from the roof a building. Just before hitting the ground, the heavier marble has

dv

dv

dx

v

dx

v dv

v22

v12

v22

v12

Work/Kinetic Energy Theorem for a Variable ForceF

dv

F

dx

dv

dv

dv

v

(chain rule)

dx

=

=

dx

dt

dx

dt

Power dropped to the ground from the roof a building. Just before hitting the ground, the heavier marble has

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

PAVG = W/Dt Average Power

P = dW/dt Instantaneous Power

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

1 W = 1 J/s

The force on a particle of mass dropped to the ground from the roof a building. Just before hitting the ground, the heavier marble hasm is given by

Choose the correct statement:

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.

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.

The average power is the same as the instantaneous power.

None of the above are correct.

Spring unstretched dropped to the ground from the roof a building. Just before hitting the ground, the heavier marble has

Person pulling

Person pushing

Fs

Fp

Fs

Fp

Work done by a Springx=0

*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 unstretched dropped to the ground from the roof a building. Just before hitting the ground, the heavier marble has

Person pulling

Person pushing

Fs

Fp

Fs

Fp

Hooke’s law

Work done by a Springx=0

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

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

1-D Variable Force Example: Spring dropped to the ground from the roof a building. Just before hitting the ground, the heavier marble has

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

k =spring constant

F(x)

x1

x2

x

relaxed position

-kx

F= - k x1

F= - k x2

1-D Variable Force Example: Spring dropped to the ground from the roof a building. Just before hitting the ground, the heavier marble has

*The work done by the spring Wsduring a displacement from x1to x2 is the area under the F(x) vs x plot between x1and x2.

F(x)

x1

x2

x

Ws

relaxed position

-kx

F= - k x1

F= - k x2

1-D Variable Force Example: Spring dropped to the ground from the roof a building. Just before hitting the ground, the heavier marble has

*The work done by the spring Wsduring a displacement from x1to x2 is the area under the F(x) vs x plot between x1and x2.

F(x)

x1

x2

x

Ws

-kx

x dropped to the ground from the roof a building. Just before hitting the ground, the heavier marble has

v1

m1

m1

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.

x dropped to the ground from the roof a building. Just before hitting the ground, the heavier marble has1

v1

m1

m1

Work - EnergyUse the fact that WNET = DK.

In this case

WNET =WSPRING = -1/2 kx2and K = -1/2 mv2

so kx2 = mv2

In the case of x1

(a) dropped to the ground from the roof a building. Just before hitting the ground, the heavier marble has

(b)

(c)

x2

v2

m2

m2

Work - EnergyIf the initialspeed of the box were doubled and its mass were halved, how far x2 would the spring compress ?

x dropped to the ground from the roof a building. Just before hitting the ground, the heavier marble has2

v2

m2

m2

Work - EnergyIf the initialspeed of the box were doubled and its mass were halved, how far x2 would the spring compress ?

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

A 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 x2.

ExampleA 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?

Example: Compressed SpringN

Fs=kx

x=0

x=-11

mg

Example: Compressed Spring 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?

The 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:

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