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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: ->->->->->->->->->->->-> scussion:
Minimum tension is zero or very close to zero Newtons. Maximum is 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: ->->->->->->->->->->->-> scussion:
`dPE = `dW_con_on = 3N *.02meters = .06J
PE = .06J
???Not sure if this is correct or if I should be using average tension, which would give me .03J???
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Your calculation shows maximum tension * displacement.
The tension is almost always less than its maximum, so as you suspect this would not be appropriate.
The definition of work for a variable force uses average force.
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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:
If the PE is correctly .06J then the Final KE will be .06J in magnitude. I suppose in terms of direction it will be -.06J
Therefore
KE = 1/2 m * vf^2
-.06J = 1/2 * .02kg * vf^2
vf^2 *1/2 *.02kg = -.06kgm*m/s^2
vf^2 = -60m^2/s^2 This will give us an imaginary number so we'll need to use 60J and reason from there
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.06 / .01 = 6, not 60.
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vf^2 = 60J
vf = +-7.7m/s
Since the original pull of the rubber band was in the positive direction, I think we need to take the -velocity here.
That seems pretty fast.
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Good intuition. You simply got an extra power of 10 when dividing in the earlier step.
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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:
This is tough. I'm going to assume the domino is at 7.4m/s velocity as it leaves the zero altitude point.
For the sake of this problem upward is the postitive direction.
We want to know how long it takes to get to 0cm/s so how long for gravity to slow the velocity -7.4m/s
`dv = a*`dt
`dt = `dv/a = -7.4m/s / 9.8m/s^2 = .76 seconds
vAve = (7.4m/s + 0m/s )/ 2 = 3.7m/s
`ds = `vAve *`dt = 3.7m/s * .76s = 2.8 m approx
That seems pretty high.
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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:
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Your extra factor of 10 (from you division error) explain the unexpectedly high velocity and vertical displacements.
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No need to revise. Your steps were all correct, and you recognized something was amiss.
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