Phy 121
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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?
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I think that the average tension is 1.5 Newtons. I figured this out in the same manor that I would find the average velocity of an object. I assumed that the starting tension was zero and since our final tension was 3 Newtons, that by adding the two values together and dividing by two that we could find the average tension.
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How much work is required to stretch the rubber band from 8 cm to 10 cm?
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We can determine the amount of work needed to stretch the rubberband by finding the product of our average force and displacement.
I know that the displacement is 2 cm. I was not sure about finding the average work done. I thought that it was 9.8 Newtons, but as I was unsure, I did not proceed any further.
You found the correct average force, 1.5 N, in the first question.
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During the stretching process is the tension force in the direction of motion or opposite to the direction of motion?
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I think that the tension force would be similar to friction and would therefore move in a direction that opposite to the direction of motion.
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Does the tension force therefore do positive or negative work?
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I think the force from the tension when released does positive work even though the direction of work is different that the direction it is pulled in. I think this might contradict what I said earlier though.
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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?
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The tension force would do 3 Newtons worth of work on the domino. I would at least assume this. I know we are given information on the mass of the item, so I am wondering if this somehow needs to be incorporated into solving for the answer. I am wondering if I need to use this formula - m = T / (a + g) ???
If so, then I think it would be solved as follows:
.02kg = 3 N/a + 9.8m/s2
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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?
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I think I would use this to solve for this question: 1/2 m v^2. It doesnt seem exactly right though because we are not necessarily working with velocity. Is there a formula I have missed???
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At this point how fast will the domino be moving?
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To solve for the acceleration, I think I need to have an exact answer for the kinetic energy and the amount of work done on the system.
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35-40
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