course Phy 201
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09:33:01 QUESTION FROM STUDENT--Please define the differnece between Fnet and Force. See if you can answer this question.
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RESPONSE --> Fnet is the sum of all forces acting on an object. The only reason this is my guess is because any time I have ever seen ""net"" it means the ""sum of"" whatever entity, in this case, force. confidence assessment: 1
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09:33:11 ** Net force is the sum of all forces acting on an object. If you're pushing your car you are exerting a force, friction is opposing you, and the let force is the sum of the two (noting that one is positive, the other negative so you end up with net force less than the force you are exerting). Your heart rate responds to the force you are exerting and the speed with which the car is moving; the accel of the car depends on the net force. **
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RESPONSE --> I understand. self critique assessment: 3
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09:35:23 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 --> If we use the formula vf^2=vo^2+2a'ds, we see that changes in a and 'ds could affect v^2 directly. I do not know why, though, that the change in v^2 is proportional to Fnet 'ds. confidence assessment: 2
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09:42:13 ** 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 --> Oh wow, I understand how the formula was applied to show exactly how much change a'ds could cause for vf^2. As for Fnet 'ds, I sort of understand. All of this feels like it is way over my head. I am going to reread it and try to put it in my own words. I see that Fnet is proportional to a because of the formula F=m*a, because if a increases or decreases, the change will be somehow reflected proportionally in F. Now for Fnet='ds=k*change in v/^2, I still do not understand this. I understand from where the first 1/2 was derived (from the formula), but in terms of the Fnet equation, I do not understand how the ""specifically for k=mass/2"" comes into play. self critique assessment: 0
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09:43:18 How do our experimental results confirm or refute this hypothesis?
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RESPONSE --> I honestly do not remember any experiment I have performed having much, if anything, to do with this. I am unable to answer this question. confidence assessment: 0
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09:45:34 ** 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 --> I understand this I think. If a remains constant, then 'ds is directly proportional to 'ds, not meaning that they are the same, but that a change in one will reflect a proportional change in the other, which will keep the slope constant. self critique assessment: 2
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