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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: ->->->->->->->->->->->-> scussion:
minimum is some number a bit more more that zero so we'll call it zero N.
maximum is 3N
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: ->->->->->->->->->->->-> scussion:
2cm = .02m
I would say fT = 1.5N * .02m = .03J
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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: ->->->->->->->->->->->-> scussion:
Tension force is opposite. It is pulling against the direction the band is being stretched.
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Does the tension force therefore do positive or negative work?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
I would say negative work because it is against the direction of motion.
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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: ->->->->->->->->->->->-> scussion:
`dW_on = fT * `ds = 1.5N *.02m = .03 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: ->->->->->->->->->->->-> scussion:
The kinetic energy of the domino would be .03J
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At this point how fast will the domino be moving?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
KE = 1/2m*vf^2
.03kg*m^2/s^2= 1/2 *.02kg *vf^2
.03kg*m^2/s^2 = .01kg * vf^2
3m^2/s^2 = vf^2
+-1.7m/s = vf
vf= +-1.7m/s
If the direction of motion was the direction we pulled the rubber band we would use the negative value here.
vf = - 1.7m/s
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&#Very good responses. Let me know if you have questions. &#