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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.
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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: ->->->->->->->->->->->-> :
3 Newtons = K * 0.1 m
K = 3 Newtons / 0.1 m
K = 30 N/m
Tmin = 30 N/m * 0.08 m = 2.4 Newtons
Tmax = 30 N/m * 0.1 m = 3 Newtons
Tave = 3 + 2.4 Newtons / 2 = 2.7 Newtons
The tension doesn't increase while the length of the rubber band increases by .1 m. There is no tension if the length is less than .08 m. So the tension increases from 0 to 30 N over a length interval of .02 m.
The value of k is therefore 3 N / (.02 m) = 150 N / m.
If you are using F = k x, then since no force is exerted for lengths less than .08 meters, x cannot be the length of the rubber band. x is instead the distance beyond the 'equilibrium length' of .08 meters. If L is the length of the rubber band, in other words, x = L - .08 meters.
The tension force function would therefore be
F = k x = k * (L - .08 m).
Extended discussion, including some graphs, is provided at the end of the linked document.
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• How much work is required to stretch the rubber band from 8 cm to 10 cm?
answer/question/discussion: ->->->->->->->->->->->-> :
W = F * D
W = 2.7 Newtons * 2 cm (0.02 M)
W = 0.054 N/m
when you multiply N by m you get N * m, not N / m
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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: ->->->->->->->->->->->-> :
Positive work done exemplifies tension is done in the direction of motion.
you might do positive work to stretch the rubber band, but that doesn't mean the tension force does positive work
you have to specify the directions then interpret the resulting conclusions
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• Does the tension force therefore do positive or negative work?
answer/question/discussion: ->->->->->->->->->->->-> :
The tension force does positive work done.
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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: ->->->->->->->->->->->-> :
Force = Ti – Tf
Force = 3 – 2.4 Newtons = 0.6 Newtons
W = F * D
W = 0.6 Newtons * 0.02 m = 0.012 N/m
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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: ->->->->->->->->->->->-> :
W = KE
KE = 0.012 N/m
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• At this point how fast will the domino be moving?
answer/question/discussion: ->->->->->->->->->->->-> :
KE = ½ mv^2
0.012 N/m = ½(.02 kg)(v^2)
V^2 = 1.2
V = + - 1.10
V = 1.1 m/s
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1 hr
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10/15 5pm
Overall this is very good, but you made an incorrect assumption which threw your first result off, and you weren't careful about directions when figuring out whether the work done by the tension force was positive or negative.
Be sure you understand the discussion at the link. No revision is necessary as long as you do understand, as I believe you will.
&#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. &#