phy 201
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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:
8cm = 1N, 9c, = 2N , 10cm = 3N
Min = 1N max = 3N
Just before a rubber band begins to exert a force, it exerts 0 N of force. At the instant it begins to exert a force, the force begins to increase; but in an instant it hasn't had time to increase, so the force is still 0 at length 8 cm. Any length greater than 8 cm results in a force greater than 0, but at the 8 cm length the force is still 0.
So the average force would be 1.5 N.
If you were to replace your 2 N with 1.5 N and rework the following, you would get correct results, and they would be a little different than the results you obtained. You don't need to actually do that, since you understand the remainder of the process very well.
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:
Fnet * ds = W
3N * .02cm = 6J
W = PE
I used 2 cm because this was the distance stretched with a force applied.
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
PE = KE =.06J = .5mv^2
V = KE /.5mv^2
V = .06J /.5 *.020kg
V = 6m/s
If instead the rubber band is used to 'shoot' the domino straight upward, then how high will it rise?
answer/question/discussion:
V0 = 0
Vf = 6m/s
F=ma
A = F/m = 3N/.020kg
A = 150m/s
Vf^2 = v0^2 * 2a*ds
Vf ^2 - v0^2 /2a = ds
Ds = 6^2 /2*150m/s
D = .12m
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25min
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&#Good work. See my notes and let me know if you have questions. &#