cq_1_141

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Phy 121

Your 'cq_1_14.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.

** 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: ->->->->->->->->->->->-> :

min tension is 0N and max tension is 3N.

Average tension is 1.5N

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• How much work is required to stretch the rubber band from 8 cm to 10 cm?

answer/question/discussion: ->->->->->->->->->->->-> :

W=F*’ds=3N*.02m=.06J

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The force is 3 N only at the maximum length. For the rest of the .02 m interval the force is less than 3 N, so the average force is less than 3 N.

You need to do this calculation based on the average, not the maximum 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: ->->->->->->->->->->->-> :

Tension force is opposite direction of motion.

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• Does the tension force therefore do positive or negative work?

answer/question/discussion: ->->->->->->->->->->->-> :

Negative 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: ->->->->->->->->->->->-> :

Not sure about this one. 3N *.02m=.06J. Since I was given a mass of .02kg, I don’t know how to work the mass in to find work. I think the mass is already in the 3N as F=ma.

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You've already calculated the work, which depends only on the force that acts on the mass and the displacement through which it acts. The mass has nothing to do with the amount of work done on it.

The velocity required to give it an equal kinetic energy, on the other hand, does depend on the mass.

The work-kinetic energy theorem says that the work done by the net force is equal to the change in its kinetic energy.

You know the work. The mass started from rest, so the change in its kinetic energy is equal to the kinetic energy it attains.

Thus you can set the expression for kinetic energy equal to the work done. Knowing the mass, you'll easily be able to solve for the velocity.

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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: ->->->->->->->->->->->-> :

If we use F=ma, then a=F/m=3N/.02kg=150m/s^2.

vf=sqrt(v0^2+2 a ‘ds)=sqrt((0m/s)^2 + 2 * 150m/s^2 * .02m)=2.45m/s

KE=.5*m*v^2=.5*.02kg*(2.45m/s)^2=.06J

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You did this assuming uniform acceleration, which does not apply to this situation. The rubber band force, and therefore the acceleration of the system, is not constant.

The work-energy theorem does apply, and gives the result you would get by assuming a force equal to the average force and proceeding as you did. So your analysis confirms the work-energy theorem for this case (or would, had you used the average rather than the maximum force).

It's important, in any case, to be able to apply the work-energy theorem without resorting to the equations of motion.

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• At this point how fast will the domino be moving?

answer/question/discussion: ->->->->->->->->->->->-> :

2.45m/s

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You haven't applied the work-energy picture to this problem, and you also found but then failed to use the average force as a basis for your results.

It shouldn't take you long to make the necessary revisions.

&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).

Be sure to include the entire document, including my notes.

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