course PHY201 p|ك|ĤXxassignment #
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12:26:56 QUESTION FROM STUDENT--Please define the differnece between Fnet and Force. See if you can answer this question.
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RESPONSE --> I do not have a solid understanding of the difference. confidence assessment: 0
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12:29:50 ** 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 --> OK I understand this. I have been ""roadblocked"" on this question for a while, trying to find a good answer, but I I eventually gave in and just admitted to not knowing what the answer was. I do understand after reading the explanation. self critique assessment: 2
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12:40:36 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 --> In terms of the concept of `ds being the area under a trapezoid where the slope of the top line of the trapezoid is represented by a, it would follow that the change in v^2 would be related to the slope a of the trapezoid over the time which is represented on the x axis. With regards to Fnet `ds, it is logical to think that Fnet and a are very closely related, such that substituting Fnet for a would give a result that may not numerically match the result that a would give, but will at least yield roughly the same slope as a. confidence assessment: 1
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12:47:17 ** 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 --> I don't think my answer was even really in the ballpark, but I copied the given answer to my notes, and I will revisit it while I study. The answer provided makes sense to me when I read through it, but I think I will need to spend some more time studying before I will really understand it. self critique assessment: 2
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12:50:41 How do our experimental results confirm or refute this hypothesis?
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RESPONSE --> The experimental results tend to show that the acceleration of an object is proportional to the amount of weight in an opposite direction (hanging via a pulley for instance) it takes to offset the motion of that object. So I would say that the results would confirm the hypothesis. confidence assessment: 1
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12:57:14 ** 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 think I understand most of this. I am a little confused though, because I feel like I went from a very solid understanding, to being kind of lost. I haven't felt comfortable with any of the questions in this assignment query. self critique assessment: 2
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12:57:16 ** 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:
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