Phy 201
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.
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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:
since the force starts at 8cm and ends at 10 cm and we have a total of 3newtons and tension increases steadily then we can say 8cm=1newton, 9cm=2newtons, and 10cm=3newtons. If we take the total newtons of 3 and divde by 2 we get Fave of 1.5newtons
How much work is required to stretch the rubber band from 8 cm to 10 cm?
answer/question/discussion:
Since we know work done is = to Fave*distance we can find how much work is done beteen 8 and 10cm. 1.5newtons*8cm=12newton/cm,1.5newtons*9cm=13.5n/cm,1.5newtons*10cm=15n/cm. If we subtract 15n/cm-12n/cm=3n/cm
Your answer is correct (except for the units, which are N * cm and not N / cm--there is no division involved. However your result could have been obtained more simply:
Work is the product of the force in the direction of the displacement and the displacement.
To stretch the rubber band requires a displacement of the one or both ends, with stretching forces exerted in the directions of any displacements. So to stretch the rubber band requires positive work.
The total displacement is 10 cm - 8 cm = 2 cm.
The force changes linearly with displacement, from 0 N to 3 N, so the average force is (0 N + 3 N) / 2 = 1.5 N.
The work is therefore force * displacement = 1.5 N * 2 cm = 3.0 N * cm.
Note that it is possible that both ends of the rubber band displace in the stretching process. The tension at each end exerts a force which is directed opposite to its displacement, so the force exerted on the rubber band at each end is in the direction of the displacement at that end. The average force exerted at the ends is still 1.5 N, and the sum of the displacements of the two ends is 2 cm (for example if one end displaces 1 cm the other will displace 1 cm ; if one end displaces .5 cm the other will displace 1.5 cm), so the sum of the work done on the two ends is still 3.0 N * cm.
During the stretching process is the tension force in the direction of motion or opposite to the direction of motion?
answer/question/discussion:
opposite the motion because the rubberband is pulling against the whatever is pulling it
Does the tension force therefore do positive or negative work?
answer/question/discussion:
negative work since it is opposite motion
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:
Since this is conservative then the work done to pull the rubberband will be equal to the work done on the domino. It will be 3newton/cm
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:
Since it potential energy is 3newtons/cm or.030joules and when potential energy decreases kinetic energy increases then the kinetic energy will be .030joules
At this point how fast will the domino be moving?
answer/question/discussion:
Since we know mass and KE and we need to find v we can use the equation KE=1/2m*v^2. If we rearrange this to find velocity we have v=sqrt((2*KE)/m). Fill in the equation v=sqrt((2*.030joules)/.02kg) we get,b=sqrt(.06joules/.02),v==sqrt(3joules/kg)v=1.7m/s
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20min
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&#Good responses. See my notes and let me know if you have questions. &#