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Work Done by Varying Forces - General Physics I - Lecture Slides, Slides of Physics

Following points are the summary of these Lecture Slides : Work Done by Varying Forces, Variable Force, Object, acting, Displacement, Versus, Variable Force, Energy Stored, Potential, Spring Constant

Typology: Slides

2012/2013

Uploaded on 07/26/2013

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Work Done by Varying Forces
Recall: the work done
by a variable force
acting on an object that
undergoes a
displacement is equal to
the area under the
graph of F versus x
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Download Work Done by Varying Forces - General Physics I - Lecture Slides and more Slides Physics in PDF only on Docsity!

Work Done by Varying Forces

  • Recall: the work done

by a variable force

acting on an object that

undergoes a

displacement is equal to

the area under the

graph of F versus x

5

x (m)

Fx (N)

F 3

F 2

F 1

x 1 x 2 x 3

The work done by F 1 is ( 0 ) 1 1 1

W = F x

Example: What is the work done by the variable force shown below?

The net work is then W 1 +W 2 +W3.

The work done by F 2 is ( ) 2 2 2 1

W = F xx

The work done by F 3 is (^) ( ) 3 3 3 2

W = F x − x

7

Example: (a) If forces of 5.0 N applied to each end of a spring cause the spring

to stretch 3.5 cm from its relaxed length, how far does a force of 7.0 N cause the

same spring to stretch? (b) What is the spring constant of this spring?

(a) For springs F∝x. This allows us to write.

2

1

2

1

x

x

F

F

Solving for x 2 : (^) ( 3. 5 cm) 4. 9 cm.

5.0 N

  1. 0 N

1

1

2 2 ⎟ =

⎠

⎞ ⎜

⎝

⎛ = x =

F

F x

  1. 43 N/cm.

3.5 cm

5. 0 N

1

1 = = =

x

F

k

(b) What is the spring constant of this spring? Use Hooke’s law:

  1. 43 N/cm.

4.9 cm

7. 0 N

2

2 = = =

x

F

Or k

8

Example: An ideal spring has k = 20.0 N/m. What is the amount of work done

(by an external agent) to stretch the spring 0.40 m from its relaxed length?

Fx (N)

x (m) x 1 =0.4 m

kx 1

( )( ) ( 20. 0 N/m)( 0. 4 m) 1. 6 J

2

1

2

1

2

1

Area under curve

2 2 1 1 1 = = = =

=

kx x kx

W

Conservation of Energy including a Spring

  • If needed, the PE of the spring is added to both sides of

the conservation of energy equation

g s i g s f

( KE + PE + PE ) = ( KE + PE + PE )

11

Example: A box of mass 0.25 kg slides along a horizontal, frictionless surface

with a speed of 3.0 m/s. The box encounters a spring with k = 200 N/m. How

far is the spring compressed when the box is brought to rest?

  1. 11 m

2 2

v

k

m x

mv kx

U K U K

E E

i i f f

i f

Given:

m = 0.25 kg

k = 200 N/m

vi = 3.0 m/s

vf = 0 m/s

Find:

x =?

Idea: we are given velocity, mass and spring constant.

Let’s use conservation of energy: kinetic energy pf the

box was transformed into elastic potential energy of the

spring.

Nonconservative Forces and Energy

  • In equation form:
  • The energy can either cross a boundary or the

energy is transformed into a form not yet

accounted for

  • Friction is an example of a nonconservative force

nc f f i i

nc f i i f

W KE PE KE PE

W KE KE PE PE or

Transferring Energy

  • By Work
    • By applying a force
    • Produces a displacement

of the system

Transferring Energy

  • Mechanical Waves
    • a disturbance propagates

through a medium

  • Examples include sound,

water, seismic

Transferring Energy

  • Electrical transmission
    • transfer by means of

electrical current

Problem Solving with Nonconservative

Forces

  • Define the system
  • Write expressions for the total initial and final energies
  • Set the W nc equal to the difference between the final and

initial total energy

  • Follow the general rules for solving Conservation of

Energy problems

Power

  • Often also interested in the rate at which the

energy transfer takes place

  • Power is defined as this rate of energy transfer

    • SI units are Watts (W)

F v

t

W

P = =

2

2

s

kg m

s

J W

= =

22

Example: A race car with a mass of 500.0 kg completes a quarter-mile (402 m)

race in a time of 4.2 s starting from rest. The car’s final speed is 125 m/s.

What is the engine’s average power output? Neglect friction and air resistance.

  1. 3 10 watts

5

2

av

= ×

t

mv

t

K

t

U K

t

E

P

f

Given:

m=500.0 kg

s = 402 m

t = 4.2 s

vf = 125 m/s

vi = 0 m/s

Find:

P =?

Idea: to compute power we need energy and time. Time

is given, while energy (kinetic) can be computed

Notice that the distance information was not needed.

Center of Mass

  • The point in the body at which all the mass may be

considered to be concentrated

  • When using mechanical energy, the change in potential energy is

related to the change in height of the center of mass