Physics 110 Problems - My Solutions
Dr. Hulan E. Jack Jr.
Chapter6 Q27
Griffith, W. Thomas; The physics of everyday phenomena: a conceptual introduction for
physics;4th Edition ISBN 0-07-250977-5
THE PROBLEM STATEMENT
Ch6 Q27. A spring gun is loaded with a rubber dart, the gun is cocked, and then fired at a target
on the ceiling. Describe the energy transformations that take place in this process.
YOU TRY IT HERE FIRST !!!
Page 1 of 3
Physics 110 Problems - My Solutions
Dr. Hulan E. Jack Jr.
Ch6 Q27. A spring gun is loaded with a rubber dart, the gun is cocked, and then fired at a target
on the ceiling. Describe the energy transformations that take place in this process.
Definitions, concepts , principles and
Energy is the ability to do work
Conservation of Energy?
What does this look like?
Discussion
This is too much !!!!
Let’s see.
At 1 the spring is completely compressed all ____.
At 3 the dart has maximum speed.
At 4 the dart has stopped at the top.
Oh , 2 is just between 1 and 3.
Not so tough after all.
Just explanations. no calculations, except for extra credit!!!
Page 2 of 3
Physics 110 Problems - My Solutions
Dr. Hulan E. Jack Jr.
Basic Solution (Minimum Expected from the student)
Ch6 Q27. A spring gun is loaded with a rubber dart, the gun is cocked, and then
fired at a target on the ceiling. Describe the energy transformations that take place
in this process.
1.
2.
3.
First the spring expands as it decompresses. The potential energy due to
compression of the spring is transformed to kinetic energy of the dart.
Just as the dart leaves the uncompressed spring it has all kinetic energy and
heads upward towards the ceiling.
Now as it rises towards the ceiling, its kinetic energy is transformed to
potentials energy.
BETTER - equations preferred but not as much verbage needed.
Energy is conserved. At each position, PE + KE = constant . (PE is potential
energy and KE kinetic energy.
1.
The spring is initially compress a maximum x , x maximum , with initial PE =
PE maximum = ½ kx maximum2 and KE = 0 because the dart is initially at rest.
As the spring decompresses, Conservation of Energy gives
PE spring + KE dart = constant ,
½ kx2 + ½ mv2 = constant ,
where k is the spring stiffness, x the compression of the spring, m the mass of the
dart, and v the velocity of the dart.
2.
Just as the dart leaves the uncompressed spring and heads upward towards
the ceiling it has all kinetic energy ½ mvmaximum2 . The value of the KE maximum
= ½ kxmaximum2 , where xmaximum is the total.
3.
Now as it rises towards the ceiling, its kinetic energy is transformed to
potentials energy as given by Conservation of Energy,
PE graviry + KE dart = constant ,
mgh + ½ mv2 = constant,
where h is the height above the spring gun.
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