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PHY 121
Your 'cq_1_15.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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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: ->->->->->->->->->->->-> n:
If it begins to exert force at 8cm, then the tension is 0N. At 10cm the tension is 3N and the average would be 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: ->->->->->->->->->->->-> n:
PE = F_ave * `ds
PE= 1.5N * 2cm
PE = 3N*cm
PE = .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: ->->->->->->->->->->->-> n:
.03J = 1/2(m*v^2)
.03J = 1/2(.02kg * v^2)
.06J = .02kg*v^2
3J/kg = v^2
V = +-1.73m/sec
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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: ->->->->->->->->->->->-> n:
v0 = 1.73m/sec
vf = 0m/sec
a = 9.8m/sec^2
F = m*g
F = .02kg * 9.8m/sec^2
F = .196kg*m/sec^2
F = .196N
`ds = .03J/(.02kg * 9.8m/sec^2)
=.03J/.196kg*m/sec^2
=.03N*m/.196N
=.153m
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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:
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*#&!
&#Your work looks very good. Let me know if you have any questions. &#