Transcript Work

Work
Work

When a force is done on an
object, energy is transferred to
the object.

The amount of energy transferred
is called work.
Work

If you apply a 200.-Newton force
to move a box 50.0 meters, how
much work have you done?
Work
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
Symbol: W
SI Unit: Joule
Equation: W = Fd
Force must be in same direction
as movement.
 Work only happens if a force
causes movement.

Is it Work?

You push against the wall. The
wall stubbornly refuses to move.


Is it Work?
No! Work is only done if a force
causes movement.
Is it Work?

An apple falls from a tree to the
ground.


Is it Work?
Yes! Because the force of
gravity causes the apple to
move, work is done on the
apple.
Is it Work?

You carry a tray above your
head while walking across the
room at constant speed.

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Is it work?
No! Although you are applying a
force to the tray, it isn’t in the
same direction that the tray is
moving.
Is it Work?

You push a lawnmower across
the lawn.

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Is it Work?
Yes! Only the horizontal
component of your applied force
does work on the lawnmower,
however.
Work or Not?

An easy way to tell if work is
being done:
Is a force acting on an object to
cause a displacement?
 Is the amount of energy an object
has (either kinetic or potential)
changing?
 If so, then work is being done!

Practice Problems

A 50.0-kg crate is lifted 25.0 m
upward by a force of 1000. N.
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How much work is being done by
the applied force?
W = Fd
W = (1000. N)(25.0 m)
W = 2.50x104 J
Practice Problems

A satellite orbits the Earth in a
circular orbit of radius = 10,000 km.
How much work does the Earth do
on the satellite?

Surprisingly, none!

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The Earth does apply a force on the
satellite, but it causes no displacement
toward or away from the Earth.
Also, neither the satellite’s kinetic nor its
potential energy changes.
Work-Energy Theorem

A net force causes acceleration.

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A net force acting over a
distance does work.
Sooo...

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If speed changes, KE changes.
Fnetd = K
Or,

W = K
Practice Problems

How much work is done in
accelerating a 500-kg vehicle
from 0 m/s to 20 m/s?

W = K
Ko = 0 J
 Kf = ½ mv2 = 100,000 J


Work done = 100,000 J
Practice Problems

What force is required to accelerate
a 0.20-kg baseball from 0 m/s to 50
m/s over a distance of 1.25 m?
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W = K
Fd = Kf – Ko
Fd = ½ mvf2 – ½ mvo2
F(1.25 m) = 250 J – 0 J
F(1.25 m) = 250 J
F = 200 N (about 45 lbs of force)
Practice Problems

Just for fun (this is fun, right?),
let’s solve the previous problem
without using our knowledge of
work or kinetic energy.

Should we get the same answer?
Practice Problems

What force is required to
accelerate a 0.20-kg baseball
from 0 m/s to 50 m/s over a
distance of 1.25 m?
F = ma
 Don’t know a, but we do know
that:

vo = 0 m/s
 vf = 50 m/s
 x = 1.25 m
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Practice Problems

vf2 = vo2 + 2ax
(50 m/s)2 = (0 m/s)2 + 2a(1.25 m)
 2500 m2/s2 = (2.50 m)a
 a = 1000 m/s2

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F = ma
F = (0.20 kg)(1000 m/s2)
 F = 200 N

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Which method do you prefer?
Practice Problems

A force of 2600 N is applied at an
angle 60º below the horizontal
against a 200.-kg crate that is initially
at rest on a frictionless surface.
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The crate moves 10.0 meters.
What is the crate’s velocity when it has
moved 10.0 meters?
vf = ???
60º
10.0 m
Practice Problems
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First, work out the horizontal
component of the force.
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The only component that does work.
Fhoriz = F*cos
Fhoriz = (2600 N)(cos60º)
Fhoriz = 1300 N
vf = ???
60º
1300 N
10.0 m
Practice Problems
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W = Fd = K
(1300 N)(10.0 m) = 13,000 J
K = 13,000 J
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Kf = 13,000 J
13,000 J = ½ mvf2
13,000 J = ½ (200 kg)vf2
vf2 = 130 m2/s2
vf = 11.4 m/s
vf = ???
60º
1300 N
10.0 m