course Phy 201 10-03-2007?y??????J??????assignment #008
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13:57:26 QUESTION FROM STUDENT--Please define the differnece between Fnet and Force. See if you can answer this question.
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RESPONSE --> The net force is the force of all things working on something Force is the push or pull on an object. confidence assessment: 1
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14:07:32 In terms of the equations of motion why do we expect that a * `ds is proportional to the change in v^2, and why do we then expect that the change in v^2 is proportional to Fnet `ds?
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RESPONSE --> fnet=mass times acceleration average acceleration times`dt=`dv vf^2=v0^2 + 2a`ds vf^2-v0^2=2a`ds vf^2-v0^2 is proportional to Fnet `ds Fnet=ma F/m=a confidence assessment: 1
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14:11:32 ** It's very important in physics to be able to think in terms of proportionality. To say that y is proportional to x is to say that for some k, y = k x. That is, y is a constant multiple of x. To say that a * `ds is proportional to the change in v^2 is to say that for some k, a * `ds = k * ( change in v^2)--i.e., that a * `ds is a constant multiple of the change in v^2. In terms of the equations of motion, we know that vf^2 = v0^2 + 2 a `ds so a `ds = 1/2 (vf^2 - v0^2), which is 1/2 the change in v^2. So a `ds is a constant multiple (1/2) of the change in v^2. Formally we have a `ds = k ( change in v^2) for k = 1/2. Now since Fnet = m a we see that Fnet is proportional to a for a given mass m, and it follows that Fnet `ds = k * change in v^2, for the appropriate k (specifically for k = mass / 2. **
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RESPONSE --> vf^2=v0^2 +2a`ds a `ds= .5(vf^2-v0^2) `ds times a is a constant multiple (.5) of vf^2=v0^2. fnet=ma Am confused: fnet/m=a Why do you use k? self critique assessment: 1
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14:12:03 How do our experimental results confirm or refute this hypothesis?
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RESPONSE --> What experimental results? From which lab? confidence assessment: 1
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14:12:52 ** We didn't actually do this part of the experiment, but on a ramp with fixed slope a `ds is simply proportional to `ds. When we measured `dt for different distances from rest down the same ramp, we were then able to determine the average and final velocities. The change in v^2 for each timing would be from 0 to vf^2. The change would therefore be just vf^2. If a `ds is proportional to the change in vf^2 then a graph of vf^2 vs. a `ds should be linear. Since a is constant we don't even need to determine it--a graph of vf^2 vs. `ds would be linear. This would confirm the hypothesis. **
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RESPONSE --> self critique assessment: 3
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