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Phy 121
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** CQ_1_15.1_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?
answer/question/discussion: ->->->->->->->->->->->-> :
Min tension is at 8cm and is 0N
Max tension is at 10cm and is 3N
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• Assuming that the tension in the rubber band is 100% conservative (which is not actually the case) what is its elastic potential energy at the 10 cm length?
answer/question/discussion: ->->->->->->->->->->->-> :
At 10cm F=3N and ‘ds =.02m
‘dW=F*’ds=3N*.02m=.06J
Since the system is not moving at this point, all energy is PE. So, PE=.06J
I believe the formula is W_c= -‘dPE. I haven’t determined a direction of movement. So I leave .06J as positive. I’m not sure that’s correct.
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The 3 N tension occurs only at one point. At every other point the tension is less than 3 N.
So you would not calculate work as if you had a constant 3 N force over the 2-cm displacement.
What is the average force over the 2 cm displacement, and what therefore is the work done by this force?
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• If all this potential energy is transferred to the kinetic energy of an initially stationary 20 gram domino, what will be the velocity of the domino?
answer/question/discussion: ->->->->->->->->->->->-> :
We let PE=KE for this transfer of energy. So, KE=.06J
Now we solve for velocity:
KE=.5*m*v^2
v=sqrt((KE)/(.5*m))=sqrt((.06J)/(.5*.02kg))=2.55m/s
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• If instead the rubber band is used to 'shoot' the domino straight upward, then how high will it rise?
answer/question/discussion: ->->->->->->->->->->->-> :
Using all PE to project the domino upward we have:
PE=m*g*y
y=(PE)/(m*g)=(.06J)/(.02kg*9.8m/s^2)=.31m
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For University Physics students:
Why does it make sense to say that the PE change is equal to the integral of the force vs. position function over an appropriate interval, and what is the appropriate interval?
answer/question/discussion: ->->->->->->->->->->->-> :
n/a Phy121
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30 min
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Good solutions, except that you miscalculated the work done by the rubber band.
Modify only that part. It's clear you know how to answer the other two questions and could easily do so with your corrected value for the energy.
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