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PHY 201
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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SEED Question 15.1
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?
answer/question/discussion: ->->->->->->->->->->->-> :
Given: m = 20 g =.02 kg
x = 2 cm = .02 m
F = 3 N
Tmin = 0, between 0 thru 8 cm
Tmax = 3 N, between 8cm thru 10 cm
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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: ->->->->->->->->->->->-> :
PE = ˝ kx^2
Hooke’s Law to find (k), k = F / x = 3 N / .02 m = 150 N/m
PE = ˝ * 150 N/m* (.02)^2 =
PE = .03 J
@& Very good.
You could also have obtained this by multiplying the average force (0 + 6 N) / 2 = 3 N by the displacement .02 m, obtaining 3 N * .02 m = .06 J.*@
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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: ->->->->->->->->->->->-> :
20 g = .020 kg
KE = PE = ˝ mv^2, solve for velocity
v^2 = 2 * PE / m
v = sqrt (2 * .03 J / .020 kg)
v = sqrt (3)
v = 1.73 m/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: ->->->->->->->->->->->-> :
PE = mgh = formula for PE
h = PE / mg
h = .03 J / (.02 m * 9.8 m/s^2)
h = .153 m = 15.3 cm
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45 mins
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SEED Question 15.1 submitted 5 Mar 11 around 9:49 PM.
@& Very well done. Do see my one note.*@