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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 **
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: ->->->->->->->->->->->-> :
(3N/10cm)=(xN/8cm)
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The tension is not proportional to the length of the rubber band, which does not begin exerting tension until its length reached 8 cm. So this proportionality is not relevant.
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x=2.4N
Average tension= (2.4+3)/2=2.7N
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How much work is required to stretch the rubber band from 8 cm to 10 cm?
answer/question/discussion: ->->->->->->->->->->->-> :
3N-2.4N=.7N
.7N*.02M=.014J
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According to the tensions you are using here the tension is always between 2.4 and 3 N. So assuming an average tension of .7 N is not appropriate to the tensions you are assuming.
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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 is resisting the direction of motion.
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Does the tension force therefore do positive or negative work?
answer/question/discussion: ->->->->->->->->->->->-> :
The tension is doing negative work on the rubber band
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The tension is the force exerted by the rubber band. Tension does not exert a force on the rubber band.
The tension exerts a force on whatever is pulling the rubber band. That force is in the direction opposite displacement, and so is 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: ->->->->->->->->->->->-> :
2.7N*2cm=5.4N
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Based on your 2.4 N and 3 N tensions, the average tension would be 2.7 N so this calculation would follow. However the units of the result would be N * cm.
The 3 N force is appropriate. The 2.4 N force is not, per my first note.
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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=M*a
5.4N=.02kg*a
a=270cm/s^2
KE=Fnet*ds
KE=5.4N*.02m
KE=.108J
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You could answer this based on the 2.7 N average force you obtained previoiusly. That isn't the correct average force, but if it was you could multiply it by the .02 m to get F_net * `ds, which would be equal to the KE, and ther rest of your calculation would follow.
However the 5.4 N came from multiplying a 2.7 N force by 2 cm, which wouldn't give you force in N. The 2.7 N itself would be the average force, and that's what you would use here.
You will still want to correct this by finding the correct average force, but the rest of your reasoning is sound
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At this point how fast will the domino be moving?
answer/question/discussion: ->->->->->->->->->->->-> :
KE=(1/2)*M*vf^2-(1/2)*M*v0^2
.108J=(1/2)*.02kg*vf^2-(1/2)*.02kg*0cm/s
vf=3.29m/s
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[ extended discussion of T vs. L and T vs. x including graphs at linked document to be provided ]
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25min
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See my notes.
It tells you in the statement of the problem that the rubber band doesn't begin exerting a force until its length is 8 cm. So for length increasing from 8 cm to 10 cm the force increases from 0 N to 3 N, not from 2.4 N to 3 N.
You shouldn't find it difficult to revise your results, most of which follow a valid reasoning process.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
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