PHY201
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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:
min = 0N
max = 3N
average = (0+3)/2 = 1.5N
• How much work is required to stretch the rubber band from 8 cm to 10 cm?
answer/question/discussion:
W= ((0N+3N)/2) *2cm = 3Ncm
• During the stretching process is the tension force in the direction of motion or opposite to the direction of motion?
answer/question/discussion:
opposite to the direction of motion
• Does the tension force therefore do positive or negative work?
answer/question/discussion:
negative work
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:
W = 3N*2cm = 6Ncm
• 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:
W=PE=KE=6Ncm
• At this point how fast will the domino be moving?
answer/question/discussion:
KE= 1/2mv^2
V=sqrt(1/2*.02kg*6Ncm)
V=24.5cm/s
The solution for v of KE = 1/2 m v^2 is v = +- sqrt( 2 * KE / m).
The units of sqrt(1/2*.02kg*6Ncm) are sqrt( kg * N * cm) = sqrt( kg * kg m/s^2 * cm) = sqrt( kg^2 m * cm / s^2); this does not give you cm/s.
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1 hour
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Most of your work is very good. Only the last calculation needs revision.
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