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.
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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: ->->->->->->->->->->->-> :
the minimum is 0 Newtons and the max is 3 Newtons. The average tension would be 1.5 Newtons.
• How much work is required to stretch the rubber band from 8 cm to 10 cm?
answer/question/discussion: ->->->->->->->->->->->-> :
`ds = 2cm = .02 m
F * `ds = `dWnet
3 Newtons * .02m = .06 Joules
• During the stretching process is the tension force in the direction of motion or opposite to the direction of motion?
answer/question/discussion: ->->->->->->->->->->->-> :
opposite to the direction of motion
• Does the tension force therefore do positive or negative work?
answer/question/discussion: ->->->->->->->->->->->-> :
negative work
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: ->->->->->->->->->->->-> :
It exerts .06 Joules on the domino
• 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: ->->->->->->->->->->->-> :
1/2mv^2
½ / ½ (.02kg) (v)^2 = .06 Joules / ½
.02 kg / .02kg(v)^2 = .12 Joules / .02 kg
SqrtV^2 = sqrt6 m^2/s^2
Vf = 2.45m/s
At this point how fast will the domino be moving?
answer/question/discussion: ->->->->->->->->->->->-> :
v0 = 0
vf = 2.45m/s
`ds = .02m
`ds = (v0 + vf) / 2 * `dt
.02m = 1.23 m/s * `dt
`dt = .02s
`ds = v0`dt + .5a`dt^2
.02m = 0 + .5a(.02s^2)
.02m/.004s^2 = .5a (.004s^2)/ .004s^2
5m/s^2 / .5 = .5 / .5 a
A = 10m/s^2
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30 min
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Very good, but you used the maximum force, rather than the average force, to calculate your work. As a result you ended up with twice as much work as you should have. No need for revision but check the link:
&#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. &#