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PHY 201
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 **
Samantha Rogers
PHY 201
Seed 14.1
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: ->->->->->->->->->->->-> :
tAve = (0N + 3N) / 2 = 1.5N
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• How much work is required to stretch the rubber band from 8 cm to 10 cm?
answer/question/discussion: ->->->->->->->->->->->-> :
`ds = 10cm - 8cm = 2cm
fAve = 1.5N
w = f * `ds
w = 1.5N * 2cm = 3Ncm or 3J
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A Joule is not a N * cm. A Joule is a N * m. The result here is .03 J, not 3 J.
I note that you correct this below.
*@
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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: ->->->->->->->->->->->-> :
stretching force is in the direction of motion and does positive work.
tension force is exhorted by the rubber band so it is in the direction opposite motion. Therefore it does negative work.
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• Does the tension force therefore do positive or negative work?
answer/question/discussion: ->->->->->->->->->->->-> :
As said before, it does negative work.
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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 work done on return is positive, +.03J
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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: ->->->->->->->->->->->-> :
`dWnet = `dKE
`dKE = .03J
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• At this point how fast will the domino be moving?
answer/question/discussion: ->->->->->->->->->->->-> :
KE = .5 m v^2
.03J = .5 .02kg v^2
sqrt((.03J) / (.5 .02kg)) = sqrt(v^2)
v = 1.7m/s
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&#This looks good. See my notes. Let me know if you have any questions. &#