Phy 231
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.
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: The minimum tension is 0 N, and the maximum is 3 N; therefore, the average tension should be 3 N/ 2 = 1.5 N.
How much work is required to stretch the rubber band from 8 cm to 10 cm?
answer/question/discussion: 'dWnet = 3N * .02 m = 0.06 J
The force is not 3 N throughout the stretching process. The 3 N tension is only achieved at the very end of the stretch, and does not act through a displacement of .02 m.
During the stretching process is the tension force in the direction of motion or opposite to the direction of motion?
answer/question/discussion: It is in the opposite direction of the motion.
Does the tension force therefore do positive or negative work?
answer/question/discussion: Negative work equal to that of the work being exerted upon it.
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.
Again assuming that the tension force is conservative, how much work does the tension force do on the domino?
answer/question/discussion: Since we have determined that the tension force is negative to the motion we have -0.06 J done on the domino
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 would be equal to the 'dWnet; therefore, if 'Wnet = - 0.06 J, KE also equals = - 0.06 J.
`dKE is equal to the work done on the object by the net force. As the rubber band contracts, is the net force in the direction of motion or opposite the direction of motion?
Consider also: Is it possible for an object to start from rest and lost kinetic energy?
At this point how fast will the domino be moving?
answer/question/discussion: - (0.06 J) = ½ (.02 kg) (v)^2; therefore,- (600 N) = v^2, which gives us - 24.49 m/s. Please not that the negative sign I placed was intended as a reminder that the value found was to be in the opposite direction of the force acting upon the rubber band initially, but the values themselves during calculation were not negative, otherwise I would not have been able to find a real number that was the square root of the value 600.
Good observation. Consider this in relation to my previous notes.
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20 minutes
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