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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: ->->->->->->->->->->->-> :
(0N + 3 N) /2 = 1.5 N
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How much work is required to stretch the rubber band from 8 cm to 10 cm?
answer/question/discussion: ->->->->->->->->->->->-> :
W = d x f = 2 cm x 1.5 N = 3 cm N
@& Very good. 90% of students in your course miss this, using the 3 N maximum force rather than the average. And most don't get the units right either.*@
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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 will be opposite of the motion because youre pulling and the rubber band is pushing
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Does the tension force therefore do positive or negative work?
answer/question/discussion: ->->->->->->->->->->->-> :
negative work oppose to the positive work done by the motion
@& The work is negative because the tension force and the displacement are in opposite directions.*@
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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: ->->->->->->->->->->->-> :
v0 = 0m/s
1.5 N
mass .02 kg
ds = 8 cm
work = f x d
1.5 x 8 cm = 12 N cm
What do I do with the mass?
@& Good, except that the rubber band only exerts force between the 10 cm length and the 8 cm length. This is only a displacement of 2 cm. When the rubber band reaches its 8 cm length is ceases to exert any force at all.
So the result would be 1.5 N * 2 cm = 3 N cm..
The mass is not relevant to this question; it is relevant to the next. So you're thinking ahead (also very good).*@
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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: ->->->->->->->->->->->-> :
fnet = ma
1.5 = .02 a
A = 75 m/s^2
is that possible?
@& You are right that KE = 1/2 m v^2.
75 m/s^2 is the average acceleration; you could solve since the initial velocity is zero and the displacement is 2 cm, so you know v0, a and `ds.
However this is a case where you don't want to consider the details of the acceleration, etc.. The rubber band tension keeps changein so acceleration isn't uniform and you would be on shaky ground using the equations of uniformly accelerated motion.
Use the fact that the change in KE is equal to the work done by the net force. You have calculated the work, though with one error. You should see from my previous note that the work is 3 N cm, not 12 N cm.
You know that the mass was released from rest so its initial KE is zero. So whatever the change in KE, that will be the final KE.
By the work-energy theorem the work done by the net force is equal to the change in KE.
So 1/2 m vf^2 = 3 N cm.
You can solve for vf.*@
KE = ½ mv^2
KE = .5 (.02 kg ) 75 m/s^2 = 140625 kg m^2/s^4
@& 75 m/s^2 is the average acceleration. This quantity is not v and it is not v^2; it doesn't belong in this calculation.
The main point is that you need to be sure you are identifying your quantities correctly. Make sure each quantity has the right units, and that each was obtained in a valid manner.*@
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At this point how fast will the domino be moving?
answer/question/discussion: ->->->->->->->->->->->-> :
A= 75 m/s^2?
Not sure how to solve this
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20 min
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@& You have a few errors in detail, but your approach to the problems is excellent. With a little more practice I believe you'll be in good shape.*@
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&#See any notes I might have inserted into your document, and before looking at the link below see if you can modify your solutions. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your old and new solutions with the expanded discussion at the link
Solution
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