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.
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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.
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Between the 8 cm and 10 cm length, what are the minimum and maximum tensions?
answer/question/discussion: minimum tension at .08 m is 0 N, maximum tension at .1 m is 3 N
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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: the elastic potential energy is proportional to the square of the distance, (.02 m)^2 = .04 m^2 * 3 N = .12 J
The elastic potential energy is indeed proportional to the square of the displacement from equilibrium. However the value of the proportionality constant will depend on the situation. Your calculation assumes that the valley of the proportionality constant is 3 N, which is not the case.
The elastic potential energy achieved in the stretching process is (assuming that the force is 100% conservative, which is not actually the case for a real rubber band) equal to the work done against the tension force, which as you found in the preceding cq question to be .03 Joules.
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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 = mv^2 = sqrt(.12 J / .2 kg) = .77 m/s = vf
The mass would be .02 kg; the KE would be .03 J instead of .12 J (see previous note).
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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: v^2 / 2a = (.77 m/s)^2 / [2 * 9.8 m/s^2] = .03 m
In terms of energy, the rubber band will rise until its .03 J of KE is converted to gravitational PE.
The gravitational PE change is weight * upward displacement = m g * `dy, so that
m g * `dy = -`dKE
and `dy = - `dKe / (m g) = - (0 - .03 J) / (.02 kg * 9.8 m/s^2) = .15 m.
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25 minutes
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You have almost all of the right ideas. Be sure to see my notes, especially the last when about conversion of the initial kinetic energy to gravitational potential energy. Let me know if you have questions.