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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.
** 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: ->->->->->->->->->->->-> :
minimum is 0 N
maximum is 3 N
average tension is 1.5 N
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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_net * 'ds
= 3 N * 0.02 m
= 0.06 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: ->->->->->->->->->->->-> :
The rubberband will exert all of its energy into the domino, so the work will be the same amount of work as the tension
0.06 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: ->->->->->->->->->->->-> :
Are you asking the same question????? Basically the amount of work put into the domino is the amount released by the rubberband. KE is basically the same thing,
so the answer is 0.06 J
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At this point how fast will the domino be moving?
answer/question/discussion: ->->->->->->->->->->->-> :
W = F_net * 'ds
'ds = 0.06 J / 3N
= 0.02 m
F = m * a
3 N = 0.2 kg * a
a = 150 cm/s^2
vf^2 = v0^2 + 2 a `ds.
= sqrt(0 + 2*150*0.02)
= 2.5 m/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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20 minutes
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145pm 4/5/2012
&#Good responses. Let me know if you have questions. &#