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PHY 201
Your 'cq_1_15.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_15.1_labelMessages **
A rubber band begins exerting a tension force when its length is 8 cm. As it is stretched to a length of 10 cm its tension increases with length, more or less steadily, until at the 10 cm length the tension is 3 Newtons.
• Between the 8 cm and 10 cm length, what are the minimum and maximum tensions?
answer/question/discussion: ->->->->->->->->->->->-> :
0 N @ 8 cm and 3 N @ 10 cm
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• Assuming that the tension in the rubber band is 100% conservative (which is not actually the case) what is its elastic potential energy at the 10 cm length?
answer/question/discussion: ->->->->->->->->->->->-> :
ave_force = 1.5 N, 'ds = 0.02 cm
'dPE = ave_foce * 'dt = 1.5 * 0.02 = 0.03 J I am assuming that 'dWcons_by = 'dPE?
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We defined `dPE as being equal and opposite to the work done by the conservative force.
When the rubber band is stretched the tension, which we assume here to be conservative, acts in the direction opposite the motion of the end of the rubber band, and therefore does -.03 J of work. The PE change is equal and opposite to this, so the PE change +.03J.
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• If all this potential energy is transferred to the kinetic energy of an initially stationary 20 gram domino, what will be the velocity of the domino?
answer/question/discussion: ->->->->->->->->->->->-> :
KE = 1/2mv^2
0.03 J = 1/2(0.02 kg)v^2
v = sqrt(0.03 J / 0.01 kg) = 1.73 m/s
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• If instead the rubber band is used to 'shoot' the domino straight upward, then how high will it rise?
answer/question/discussion: ->->->->->->->->->->->-> :
the average force was calculated to be 1.5 N, but at 10 cm the the net_force was 3 N. The mass is 0.02 kg. Therefore, the the acceleration would be
F = m * a => a = F / m
3 kg*m/s^2 / 0.02 kg = 150 m/s^2 (this seems unreal because the final velocity is 1.73 m/s.
vf^2 = 2 a 'ds => 'ds = vf^2 / 2a = (1.73 m/s)^2 / (2 * 150 m/s^2) = 0.010 m. This cannot be right.
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150 m/s^2 would be the the maximum acceleration. But acceleration is not constant. The average acceleration would be 75 m/s^2.
This would give you a result of .02 m, which is the 2 cm through which the rubber band tension acts as it shortens from 10 cm to 8 cm.
As the mass rises its acceleration is 9.8 m/s^2, not 75 m/s^2. You could in fact use this acceleration, along with the 1.73 m/s initial velocity, to determine the height to which the mass would rise.
But you need to solve this problem using energy conservation.
Between release and the maximum height the .03 Joules of elastic PE will convert to .03 Joules of gravitational PE. This means that gravity will do work -.03 Joules on the rising mass.
As the mass rises, gravity exerts a downward force, so gravity in fact does negative work on the mass.
So through what distance will the mass have to rise in order for gravity to do -.03 J of work on it?
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Very good work on the first couple of questions.
You did have some errors on the last question, on which your work is worth revising.
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Be sure to include the entire document, including my notes. &#
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