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Phy 241
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: ->->->->->->->->->->->-> :
Since the tension increases steadily with the cm of stretch it should be increasing at about .3N/cm so the 8cm tension should be about 2.4N. The average tension would be about 2.7N
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The rubber band doesn't begin exerting tension until its length is 8 cm. Only length in excess of 8 cm will contribute to tension.
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• How much work is required to stretch the rubber band from 8 cm to 10 cm?
answer/question/discussion: ->->->->->->->->->->->-> :
In this case I would say it would be the average Tension force times the distance or 2.7N*2cm=5.4J
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Accepting your tension calculation, this is correctly reasoned and would be correct.
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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 would be opposite the direction of motion since it is opposing the band being stretched.
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• Does the tension force therefore do positive or negative work?
answer/question/discussion: ->->->->->->->->->->->-> :
The tension could really do either, depending on if you are counting the pull as positive or negative, but since we would likely call the pull positive we would then call the tension negative.
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Right.
More clarification:
The tension pulls in the direction opposite the motion of your finger. That would apply to either finger, or to both.
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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: ->->->->->->->->->->->-> :
Assuming the force is conservative and a perfect world of no other losses, it would do the same 5.4J 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: ->->->->->->->->->->->-> :
Fnet=`dw/`ds=5.4J/2cm=2.7N
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The kinetic energy would, by the work-kinetic energy theorem, be 5.4 Joules (if we accept your value of the average tension).
Kinetic energy and work are commensurate units. Kinetic energy and force have different units. So the answer couldn't be 2.7 N.
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• At this point how fast will the domino be moving?
answer/question/discussion: ->->->->->->->->->->->-> :
v^2=2Fnet/m=2*2.7N/.2kg=27cm/s ; v=5.2cm/s
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Right reasoning but v^2 = 2 * KE / m, not 2 * F / m. The units of your calcualtion wouldn't work out.
N / kg is (kg m / s^2) / kg = m / s^2, so sqrt(N / kg) would be m^.5 / s.
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*#&!
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You've got some errors in quantities and units, so be sure you see my notes.
You've also misinterpreted the rule for tension, though much of your subsequent reasoning is good.
Be sure to check out my notes and the link. No revision is necessary as long as you're sure you understand.
&#See any notes I might have inserted into your document, and before looking at the link below see if you can modify your solutions. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your old and new solutions with the expanded discussion at the link
Solution
Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified.If your solution is completely consistent with the given solution, you need do nothing further with this problem. &#
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