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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: ->->->->->->->->->->->-> :
min = 0 Newtons
max= 3 Newtons
average= 1.5 Newtons
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• How much work is required to stretch the rubber band from 8 cm to 10 cm?
answer/question/discussion: ->->->->->->->->->->->-> :
`ds= 2cm or .02m and average force= 1.5 newtons so .02m*1.5n= .03 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 (the rubber band wants to fly back)
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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: ->->->->->->->->->->->-> :
+0.03 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: ->->->->->->->->->->->-> :
also +0.03 Joules
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• At this point how fast will the domino be moving?
answer/question/discussion: ->->->->->->->->->->->-> :
KE = 1/2 m v^2, so v = sqrt(2 * KE / m).
v = sqrt( 2 * .03 Joules / (.02 kg) ) = sqrt(3 m^2 / s^2) = 1.7 m/s
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*#&!
&#Very good work. Let me know if you have questions. &#