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Phy 241
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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: ->->->->->->->->->->->-> :
Since tension increases steadily starting at 8cm it starts with 0N and ends with 3N at 10cm. The avg tension would be (3N+0N)/2=1.5N
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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: ->->->->->->->->->->->-> :
If the system is 100% cons we would say the PE is equal to the work required to stretch it which would be `dW=Fave*`ds=1.5N*.02m=.03J
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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: ->->->->->->->->->->->-> :
The .02kg domino would have the same amount of KE as the system had PE so KE=.5mv^2 so v^2=2KE/m=2*.03J/.02kg=3m^2/s^2 ; v=1.7m/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: ->->->->->->->->->->->-> :
The force of gravity on the domino would be 9.8m/s^2*.02kg=.196N
At its highest point, the KE the domino gets from the rubber band would be equal to the work being done by gravity. This would happen at .03J=.196N`ds ; `ds=.03J/.196N=.15m
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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: ->->->->->->->->->->->-> :
PE is equal to avg force multiplied by change in position. On a graph this is the same as saying the base times the midpoint of the line which makes a trapezoid and yields the area under the line. This is really the same thing as the integral, which gives us the function for area under a line.
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&#Very good responses. Let me know if you have questions. &#