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Phy 121
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.
** CQ_1_14.1_labelMessages **
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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 Newton and the maximum tension is 3 Newtons, therefore the average tension is 1.5 Newtons.
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• How much work is required to stretch the rubber band from 8 cm to 10 cm?
answer/question/discussion: ->->->->->->->->->->->-> :
.03 Joules of work is required to is required to stretched the rubber bands because 1.5N*.02m = .03J.
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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: ->->->->->->->->->->->-> :
During this process the tension force is in the opposite direction of motion.
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• Does the tension force therefore do positive or negative work?
answer/question/discussion: ->->->->->->->->->->->-> :
Tension force therefore, does negative work.
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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: ->->->->->->->->->->->-> :
The tension force does .03 Joules of work on the domino because 1.5N*.02m = .03J.
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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: ->->->->->->->->->->->-> :
Assuming that this is the only force acting upon the domino, the KE would be .03J because it is the same as the work done by the net force, and it was .03J because ‘dW_net_ON = ‘dKE.
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• At this point how fast will the domino be moving?
answer/question/discussion: ->->->->->->->->->->->-> :
The final velocity is the domino will be 1.73m/s because KEf = 1/2mv^2, so .03J = ½*.02kg*v^2, which is 1.73m/s.
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Excellent.
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