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phy 121
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** CQ_1_14.1_labelMessages **
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: ->->->->->->->->->->->-> :
The tension at 10cm is 3 N. A stretch of 10 cm exerts a tension force of 3 N. The work done on the rubber band is therefore 0.1 M * 3 N = 0.3J. If between 8 and 10 cm 0.3J of work is done, we can say that the difference of 2 cm corresponds to a change in work of 0.3J so the initial tension must be zero because I don’t have any other numbers to work with and my comprehension is far too poor at this point to do any more.
It wouldn’t be valid to try 0.8M * 3N because 3 N is the force at 10 cm. I don’t have a mass or another newton measurement to go on so im completely clueless.
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Not bad.
However the force reaches 3 N only at the 10 cm position, and if the length is less than 8 cm there is no force.
So the average force for the 2 cm interval is not 3 N.
If the force increases from 0 N to 3 N on a 2 cm interval, what should be the average force?
Using this average force, how much work is therefore done on this interval?
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• How much work is required to stretch the rubber band from 8 cm to 10 cm?
answer/question/discussion: ->->->->->->->->->->->-> :
0.3J
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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: ->->->->->->->->->->->-> :
The tension force is in the direction of motion
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If, for example, the rubber band is being pulled to the right, then the rubber band itself pull back to the left. So the tension force is in the direction opposite the motion.
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• Does the tension force therefore do positive or negative work?
answer/question/discussion: ->->->->->->->->->->->-> :
The tension force is therefore positive.
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The question is about the work done by the tension force.
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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: ->->->->->->->->->->->-> :
At 10 cm 0.3J of tension force is being done. If we release this then the 0.3J is transferred to the 0.02kg mass in the form of kinetic energy. We could find the velocity of this if we wanted to using that `dKE is 1/2mv^2. Not counting friction, thermal energy and stuff like that 0.3J of work is done on the domino.
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Very good (except that the 0.3 J quantity will need to be modified).
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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: ->->->->->->->->->->->-> :
The kinetic energy will then be 0.
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The rubber band was stretched to its 10 cm length, then released and allowed to snap back. When it gets to the 8 cm length it will have reached its maximum speed.
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• At this point how fast will the domino be moving?
answer/question/discussion: ->->->->->->->->->->->-> :
`dKE = 1/2mv^2
2*`dKE/m = v^2
V = `sqrt 2*(0.3J) / 0.02kg
V= 5.6m/s
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Well done, though the 0.3 J figure will need to be changed.
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You'll need to refigure and rethink a couple of things, but your overall approach is good.
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