PHY 231
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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: ->->->->->->->->->->->-> :
At 10 cm in length occurs the maximum tension of 3 Newtons. I do not know how to calculate tension, but I know at 8 cm in length, occurs the minimum tension ( which might be 0 Newtons ), notwithstanding the instance where the rubber band is undisturbed. I am guessing 2.4 Newtons, using the law of proportions ( 3 * 8 /10 ). I would think you would have to know the rubber band's initial length in order to find its real point of lowest tension force or its acceleration. Thus, I believe the average tension is 2.7 Newtons between this interval of lengths.
The rubber band begins exerting for at the 8 cm length, so presumably at this length it is exerting a force which is very close to 0.
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• How much work is required to stretch the rubber band from 8 cm to 10 cm?
answer/question/discussion: ->->->->->->->->->->->-> :
To stretch the rubber band the rubber band 2 cm, it takes .06 J or .06 N-m ( .02 m * 3 Newtons ).
That would be based on the maximum force exerted over the 2-cm interval. The result needs to be based on the average force.
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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: ->->->->->->->->->->->-> :
During the stretching process, the tension force I believe is in the opposite direction of motion, once past its modulus of elasticity because the rubber band by nature experience less stress and strain at rest. Before that point at which it exceeds its yield strength, I think the stretching process is moving in the direction of motion. Thus, anything past that point, would be in the direction opposite of motion.
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• Does the tension force therefore do positive or negative work?
answer/question/discussion: ->->->->->->->->->->->-> :
I believe the tension force therefore does negative work, after exceeding its yield strength, because it is stretching the rubber band beyond its limits, its natural state of rest, causing it to be reshaped, remolded, and physically changed. And before that point, I believe it would do positive work.
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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: ->->->->->->->->->->->-> :
I would say that the tension force is .01568 N-m or .01568 Joules ( .02 kg * 9.8 m/s^2 * .08 m ).
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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: ->->->->->->->->->->->-> :
I am unsure how to reason out its kinetic energy. I know the formula previously given is KE = 1/2 * m * (( vf )^2). I know the final velocity will not be 0 m/s. If I use the acceleration as 9.8 m/s^2, the displacement as 2 cm, and the initial velocity as 0 cm/s, I attain the final velocity of .392 m/s, which does not seem likely. However, I attain the KE of
.00153664 J, which is extremely small.
KE = 1/2( .02 kg )(( .392 m/s)^2)
= .00153664 kg m^2/s^2
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• At this point how fast will the domino be moving?
answer/question/discussion: ->->->->->->->->->->->-> :
I would assume the domino would be moving at .392 m/s. Once again, I am befuddled how to solve this problem.
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30 Minutes
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I've inserted a few notes, but you need to check the link. A revision would be in order here, since a subsequent experiment depends on a similar analysis.
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