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Phy 231
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: ->->->->->->->->->->->-> :
Min tension = 0N @ 8 cm
Max Tension = 3N @ 10 cm
ave Tension = (Tf + To) /2 = 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 to stretch = f * 'ds = 3 N * 2 cm = 6 N cm
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The maximum tension doesn't act for 2 cm, nor does the minimum tension.
You can use the average tension to get the correct result.
*@
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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: ->->->->->->->->->->->-> :
Tension force is opposite to direction of motion
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Does the tension force therefore do positive or negative work?
answer/question/discussion: ->->->->->->->->->->->-> :
The tension does negative work against the pulling force.
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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: ->->->->->->->->->->->-> :
'dW (tension on domino) = T force * 'ds = 3N * 0.02m = 0.06 N m = 0.06 J
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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 = 0.06 Nm
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At this point how fast will the domino be moving?
answer/question/discussion: ->->->->->->->->->->->-> :
KE = .5 * m*v^2
v= 'sqrt(kE / (.5*m)) = 'sqrt( 0.06 kg*m^2/s^2) / (.5 * .02 kg) = 2.45 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 mins
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@&
Good, but work should have been calculated using the average tension.
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&#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
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