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Kinetic Energy and Work
Introduction
The traditional definition of energy is the ability to do work.
Kinetic Energy is the energy due to motion.
Translational Kinetic Energy is the energy moving between two points.
Rotational Kinetic Energy is the energy due to rotation around an axis.
Vibrational Kinetic Energy is the energy due to vibration about a fixed
point.
Gas molecules and many objects have a Kinetic Energy that is the sum of
the three types of Kinetic Energy.
We will restrict our current problems to only Translational Kinetic
Energy.
v22 - v12
Wnet = Fd = mad = m ( --------------- ) d = 
mv22
- mv12
2d
KE = mv2
Wnet = /\KE
The above equations tell us that work must be done on an object to speed up the
object. If the object does work, then it slows down. This is true
because the net work on the object is equal to its change in Kinetic Energy.
Table of Contents
Problems
| 1. |
The USS North Carolina shots a 6000 kg
projectile at 3.00 x 103 km/h. What is the Kinetic Energy
of the projectile?
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| 2. |
Assuming no air resistance, how much work
must be done to accelerate a 2500. kg car from 30.0 km/h to
100.0 km/h?
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| 3. |
A 2.00 kg hammer moving at 5.00 m/s hits a
nail and comes to rest. What is the net work on the hammer?
What is the net work on the nail?
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| 4. |
A car traveling at 80.0 km/h can brake to
rest in 30.0 m. If the same force is applied to the car at 160.0
km/h, what is the stopping distance? |
Table of Contents
Answers
| 1. |
The USS North Carolina shots a 6000 kg
projectile at 3.00 x 103 km/h. What is the Kinetic Energy
of the projectile?
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3.00 x 103 km
1h 1000 m
-------------------- x ----------- x ------------- = 833 m/s
1
h
3600 s 1 km
KE = 
mv2 = x 6000 kg x
(833 m s -1)2 = 2.08 x 105 J
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| 2. |
Assuming no air resistance, how much work
must be done to accelerate a 2500. kg car from 30.0 km/h to
100.0 km/h?
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30.0 km 1
h 1000
m
100.0 km 1
h 1000 m
----------- x ------------ x ----------- = 8.33
m/s
-------------- x ------------- x ------------ = 27.8 m/s
1
h 3600
s 1
km
1
h
3600 s 1 km
W = /\KE =
mv22
- mv12
W =
x 2500. kg x (27.8 m/s)2
-
x 2500. kg x (8.33 m/s)2 = 8.79 x 10 5 J
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| 3. |
A 2.00 kg hammer moving at 5.00 m/s hits a
nail and comes to rest. What is the net work on the hammer?
What is the net work on the nail?
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W hammer = /\KE =
mv22 -
mv12
W hammer =
x 2.00 kg x (0 m/s)2
- x 2.00 kg x (5.00 m/s)2 = - 25.0 J
W nail = - W hammer
= -( -25.0 J) = 25.0 J
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| 4. |
A car traveling at 80.0 km/h can brake to
rest in 30.0 m. If the same force is applied to the car at 160.0
km/h, what is the stopping distance?
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Force and displacement are in opposite directions
while breaking.
W = - Fd = /\KE = 0 -
mv2
Fd =
mv2
Force and mass are constant.
d proportional v2
If the velocity is doubled, the stopping
distance is quadrupled.
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Table of Contents
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