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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.
** 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: ->->->->->->->->->->->-> scussion:
The minimum tension would be effectively 0 at 8cm, or just barely above zero, some miniscule fraction of a Newton. The maximum tension would be 3 N, at the 10cm
stretch.
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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: ->->->->->->->->->->->-> scussion:
If it is increasing steadily, than the average tension over the course of that 2cm stretch is 1.5N. Applied over .02m, that means that by being stretched, .03J of
work is being done to the rubber band, so its KE is decreasing by .03J. If the tension force is 100% conservative, then that .03J is being regained as PE, so its
elastic potential energy is .03J.
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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: ->->->->->->->->->->->-> scussion:
.03J of work are done to the domino, increasing its KE by .03J.
.03kg*m^2/s^2=.5(.02kg)(v^2)
v^2=3m/s
v=+-1.7m/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: ->->->->->->->->->->->-> scussion:
At the instant it is released, we can still expect it to have a velocity of 1.7m/s, as we just calculated.
After that, it is subject to the force of gravity, and begin to decelerate by -9.8m/s^2, since gravity is acting in the direction opposite motion.
v0=1.7m/s
vf=0
vf^2=v0^2+2a`ds
0=(1.7m/s)^2+2(-9.8m/s^2)(`ds)
-3m^2/s^2=-19.6m/s^2*`ds
.15m=`ds
We can expect the domino to rise about .15m.
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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: ->->->->->->->->->->->-> sion:
If force is not constant, we can look at the changing slope of a force vs. position graph to calculate that changing relationship with a function. If the change in
force vs. change in position represents the work done (because work is the product of a force and the displacement through which it is applied), then the slope of that
graph, or its integral, tells us how much work is being done at each position. Since `dPE is equal and opposite `dW_ON, the integral also tells us how much PE has
changed at each position.
In this case, the interval in question is from position 8 cm to position 10 cm, because we know the force being applied at those positions.
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20 minutes
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&#Very good responses. Let me know if you have questions. &#