PHYS 152
Download
1 / 28

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


  • 104 Views
  • Uploaded on

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.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'PHYS 152 Fall 2005 Thursday, Sep 22 Chapter 6 - Work, Kinetic Energy & Power' - royal


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
Slide1 l.jpg

PHYS 152

Fall 2005

Thursday, Sep 22

Chapter 6 - Work, Kinetic Energy & Power


Exam 1 l.jpg
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.


Slide3 l.jpg

Supplemental Instruction

Exam Review Session

Tuesday Sept. 27th

7-9pm

RHPH 172


Slide4 l.jpg

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:


Slide5 l.jpg

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


Slide6 l.jpg

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


Review constant force l.jpg

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 = Fr

  • If = 90o no work is done.

No work done by T

No work done by N


Work kinetic energy theorem l.jpg
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}


Work done by lifting l.jpg

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 = mgr= -mg r

  • Now find the work done bythe hand:

    WHAND = FHANDr= FHANDr


Work done by lifting10 l.jpg

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 Lifting

Work/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


Lifting vs lowering l.jpg

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. Lowering

Lifting

Lowering

Wg= -mg r

Wg= mg r

WHAND= FHANDr

WHAND= -FHANDr

WHAND = - Wg

WHAND = - Wg


Work done by gravity l.jpg

= dropped to the ground from the roof a building. Just before hitting the ground, the heavier marble hasFr1+ Fr2+ . . . +Frn

= F(r1+ r2+ . . .+ rn)

r1

=Fr

= Fy

r2

r

r3

Wg = mgh

rn

Work done by gravity...

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

m

mg

h

j


Work done by variable force 1d l.jpg

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 = Fx

area under F vs. x plot:

Wg

x

x

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

dW = F(x) dx.


Work kinetic energy theorem for a variable force l.jpg

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 Force

F

dv

F

dx

dv

dv

dv

v

(chain rule)

dx

=

=

dx

dt

dx

dt


Power l.jpg
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


Slide16 l.jpg

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.


Work done by a spring l.jpg

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 Spring

x=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.


Work done by a spring18 l.jpg

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 Spring

x=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 l.jpg
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 spring20 l.jpg
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 spring21 l.jpg
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


Work energy l.jpg

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

v1

m1

m1

Work - Energy

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.


Work energy23 l.jpg

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

v1

m1

m1

Work - Energy

Use 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


Work energy24 l.jpg

(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 - Energy

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


Work energy25 l.jpg

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

v2

m2

m2

Work - Energy

If 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


Example l.jpg

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.

Example


Example compressed spring l.jpg

A 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 Spring

N

Fs=kx

x=0

x=-11

mg


Example compressed spring28 l.jpg
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:


ad