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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.
** 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: ->->->->->->->->->->->-> :
‘ds = 2cm
The maximum tension would be 3N, the minimum is maybe 1N? if so average would be 2N
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The rubber band just begins exerting tension when its length is 8 cm, so its tension is 0 at that length. The tension then increases gradually to 3 N at length 10 cm.
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• How much work is required to stretch the rubber band from 8 cm to 10 cm?
answer/question/discussion: ->->->->->->->->->->->-> :
‘ds = .002m
Using my average 2N guess, 2N * .002m = .004 Joules of work
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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 itself is acting against the system, so opposite the direction of motion.
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• Does the tension force therefore do positive or negative work?
answer/question/discussion: ->->->->->->->->->->->-> :
The tension force 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: ->->->->->->->->->->->-> :
.004 Joules = .02kg * m^2/s^2
.2m^2/s^2
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A Joules is a kg m^2 / s^2. So .004 Joules is .004 kg m^2 / s^2 (actually it's .04 Joules = .04 kg m^2 / s^2, per my previous note).
Tension does -.04 Joules of work when the rubber band is being stretched. When the rubber band returns to its initial length, if the tension at every position is the same as when it was being stretched, the tension acts in the direction of displacement and does +.04 Joules of work on the domino.
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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 = ½ * .02kg * .2m^2/s^2 = .002 Joules
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If .04 Joules of work were done by the net force on a domino initially at rest, then the domino would have .04 Joules of KE.
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• At this point how fast will the domino be moving?
answer/question/discussion: ->->->->->->->->->->->-> :
all this is just guessing but the only thing I can think of is to use the .2m^2/s^2
sqrt(.2m^2/s^2) = .45m/s
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The displacement was .02 m, not .002 m, so the work is .04 J rather than .004 J. Easily corrected.
The tension at 8 cm is 0, not 1 N. Be sure you understand why (per my note).
Changing the problem so that the tension at the 8 cm length is 1 N, your solution is substantially correct, except for the results of your decimal-place error.
The only exception is your use of .02 kg m^2 / s^2, and the implications of that result.
See my notes. It's up to you whether you want to submit a revision; if you're sure you understand, it's not necessary, but if you want my feedback you're welcome to submit the revision in the usual manner, complete with &&&& before and after insertions..
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