Phy 201
Your 'cq_1_14.1' 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, and what do you think is the average tension?
answer/question/discussion: At .08 m the minimum tension should be 0 N, at .10 m the maximum tension should be 3 N, the average tension between these two would be 1.5 N
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How much work is required to stretch the rubber band from 8 cm to 10 cm?
answer/question/discussion: W = Fd = .09 m * 1.5 N = .135 J, .10 m * 3 N = .30 J, Wnet = .135 J + .30 J = .435 J
Nothing is being displaced .09 m. This is the average length of the rubber band, but the rubber band does not even begin exerting a force until it is .08 m long. Between this length and the .10 cm final length the displacement is only .02 m.
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During the stretching process is the tension force in the direction of motion or opposite to the direction of motion?
answer/question/discussion: the tension force is acting in the opposite direction as the motion
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Does the tension force therefore do positive or negative work?
answer/question/discussion: because there appears to be a force acting upon the rubber band, the rubber band will have a negative force acting against the motion
The rubber band is released and as it contracts back to its 8 cm length it exerts its tension force on a domino of mass .02 kg, which is initially at rest.
Does the tension force therefore do positive or negative work? You haven't committed to an answer to this question, although you have addressed just about everything you need to know in order to answer it.
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Again assuming that the tension force is conservative, how much work does the tension force do on the domino?
answer/question/discussion: W = 3 N * .02 m = .06 J
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Assuming this is the only force acting on the domino, what will then be its kinetic energy when the rubber band reaches its 8 cm length?
answer/question/discussion: KE = .06 J; the kinetic energy will be equal to the net work done on the object
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At this point how fast will the domino be moving?
answer/question/discussion: KE = 1/2mv^2 = .06 kg * m/s^2 * m = 2m * v^2; sqrt(.06 kg * m/s^2 * m) / 2(.02 kg) = sqrt(v^2); 1.2 m/s = v
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20 minutes
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Very good. See my first note; you have essentially corrected that error on your subsequent answers.
My second note deals with the question of whether the work done by the tension force is positive or negative. You don't have to resubmit this entire document, but you should submit a clarification of your answer to that particular question.