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PHY 121
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.
** CQ_1_15.1_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?
answer/question/discussion: ->->->->->->->->->->->-> :
Minimum tension: 0 Newtons
Maximum tension: 3 Newtons
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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: ->->->->->->->->->->->-> :
3 Newtons
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This is the force, not the potential energy.
The potential energy would be the work done to stretch the system to this length.
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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 = .5mv^2
mv^2 = 2KE
v^2 = 2KE/m
v = `sqrt 2KE/m
v = `sqrt 2*3N/.02 kg
v = `sqrt 300 m/s
v = +- 17.32 m/s
The negative answer does not make sense in this case, so it would be 17.32 m/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: ->->->->->->->->->->->-> :
v = 17.32 m/s
I am doing this because it's going to go from 0 and back to zero, so I am using this as
the average velocity as well as the change in velocity.
a = 9.8 m/s
`dt = `dv/a
`dt = 17.32 m/s / 9.8 m/s
`dt = 1.77 s
vAve = `ds/`dt
`ds = vAve * `dt
`ds = 17.32 m/s * 1.77 s
`ds = 30.7 m
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For University Physics students:
Why does it make sense to say that the PE change is equal to the integral of the force
vs. position function over an appropriate interval, and what is the appropriate
interval?
answer/question/discussion: ->->->->->->->->->->->-> :
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** **
30 min
** **
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See my note above and give it a minute's thought before going to the document linked below.
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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
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.