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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.
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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, and what do you think is the average tension?
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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How much work is required to stretch the rubber band from 8 cm to 10 cm?
answer/question/discussion: ->->->->->->->->->->->-> n:
`dW = F_ave*`ds
=1.5N*(10cm-8cm)
=3N*cm
=.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: ->->->->->->->->->->->-> n:
It would be in the opposite direction of motion because it is resisting the displacement caused by the stretching force
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Does the tension force therefore do positive or negative work?
answer/question/discussion: ->->->->->->->->->->->-> n:
negative work for the same reason
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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: ->->->->->->->->->->->-> n:
1.5N
`ds = 2cm
`dW = 1.5N*2cm
=3N*cm
I assume that this is equal to the previous answer since the rubber band is conservative
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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: ->->->->->->->->->->->-> n:
v0= 0m/sec
m = .02kg
`ds= 2cm
F=1.5N
F_net = m*a
A = F_net/m
A = 1.5N/.02kg
A = 1.5 m*kg/sec^2/.02kg
A = 75m/sec^2
Vf^2= v0^2 +2*(a*`ds)
Vf ^2= 2(75m/sec^2*.02m)
Vf^2 = 3m^2/sec^2
Vf = +-1.73m/sec
KE = 1/2(m*v^2)
KE =1/2 (.02kg)*(1.73m/sec)^2
KE = 1/2(.02kg)*(2.99m^2/sec^2)
KE = 1/2(.06N)
KE = .03N
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Good.
Your units are wrong on the last two steps, though. Should read
KE = 1/2(.06J)
and
KE = .03J
No need for a revision.
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