Phy 231
Your 'cq_1_15.1' 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?
answer/question/discussion: ->->->->->->->->->->->-> :
Minimum Tension = 0N
Maximum tension = 3N
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Assuming that the tension in the rubber band is 100% conservative (which is not actually the case) what is its elastic potential energy at the 10 cm length?
answer/question/discussion: ->->->->->->->->->->->-> :
PE = .1m * 3N = .3 J
3 N is the max force, not the ave force.
This force is exerted through the 2 cm interval from 8 cm to 10 cm, not through a 10 cm interval. The rubber band exerts no tension force if its length is shorter than another length at which the tension force is zero.
Note that if the force function is linear, it can be integrated to give a result which is equal to the result you get when you multiply average force by displacement.
Recall also that an integral of any function over an interval is equal to the average value of that function on the interval, multiplied by the length of the interval.
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If all this potential energy is transferred to the kinetic energy of an initially stationary 20 gram domino, what will be the velocity of the domino?
answer/question/discussion: ->->->->->->->->->->->-> :
KE = .3J
.3J = (1/2)(.02kg)(v^2)
V^2= 30 m^2/s^2
V = +/- 5.48m/s direction and force are positive, so vf = 5.48m/s
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If instead the rubber band is used to 'shoot' the domino straight upward, then how high will it rise?
answer/question/discussion: ->->->->->->->->->->->-> :
Fnet = 3N (9.8m/s^2 * .02kg) = 3N - .196N = 2.084N
PE = 2.084N *.1m = .02084J
.02084J = 3N * ds
Ds = .0069 m in the air = .69cm
#####I was not really sure about this part
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20 minutes
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There are some errors in your calculation of the elastic potential energy. This leads to some valuable calculus considerations.
Before looking at the link, see what you can do based on my notes. Then you can check the link for a full discussion.
See also the discussion related to the last question.
&#See any notes I might have inserted into your document. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your solutions with the expanded discussion at the link
Solution
Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified.If your solution is completely consistent with the given solution, you need do nothing further with this problem. &#