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.
• 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: ->->->->->->->->->->->-> :
Minimum Tension = 0 Newtons
Maximum Tension = 3 Newtons
Average Tension = 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: ->->->->->->->->->->->-> :
W = Force * `ds
W = 3 Newtons * .02 m
W = .06 Joules
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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: ->->->->->->->->->->->-> :
Opposite
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• Does the tension force therefore do positive or negative work?
answer/question/discussion: ->->->->->->->->->->->-> :
Negative
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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: ->->->->->->->->->->->-> :
W = Force * `ds
W = 3 Newtons * .02 m
W = .06 Joules
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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: ->->->->->->->->->->->-> :
F = m * a
3 Newtons = .02 kg * a
a = 150 m/s/s
vf^2 = v0^2 + 2 * a * `ds
vf^2 = (0 m/s)^2 + 2 * 150 m/s/s * .02 m
vf^2 = 6 m^2/s^2
vf = 2.45 m/s
KE = .5 * m* v^2
KE = .5 * .02 kg * (2.45 m/s)^2
KE = .06 Joules
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• At this point how fast will the domino be moving?
answer/question/discussion: ->->->->->->->->->->->-> :
vf = 2.45 m/s
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25 minutes
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&#This looks very good. Let me know if you have any questions. &#