course phy 121 I tried to do this query and soon realized that I was very confused. I went through the query without answering the questions 2 or 3 times, using it as a tutorial of sorts. I then went through and answered the questions using my notes. I found that I am still unclear on some of the items. I don't know if this is because of taking the 2 week break, or if the topics just haven't been covered in depth yet. ???????????assignment #008008. `query 8
......!!!!!!!!...................................
09:04:03 QUESTION FROM STUDENT--Please define the differnece between Fnet and Force. See if you can answer this question.
......!!!!!!!!...................................
RESPONSE --> Fnet is all of the forces acting on an object whether they are positive or negative. When you add them all up you get the net force. confidence assessment: 2
.................................................
......!!!!!!!!...................................
09:06:09 ** 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. **
......!!!!!!!!...................................
RESPONSE --> I understand how the acceleration depends on the net force. An object can only accel if there is force acting on it and you have to take into account all of the forces. I also understand why HR responds to the force you are exerting, however, I do not see the connection between HR and the speed with which the car is moving. self critique assessment: 2
.................................................
......!!!!!!!!...................................
09:16:20 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?
......!!!!!!!!...................................
RESPONSE --> This has to do with the equation vf^2 = v0^2 + 2a*ds If you rearrange it a*ds = 1/2 (vf^2 - v0^2) Where vf^2 - v02 is the change in v^2 So a*ds is constantly 1/2 of the change in v^2 The second part I am sketchy on. If the change in v^2 is proportional to Fnet*ds then the equation looks something like this Fnet*ds = C * dv^2 where C is a constant Fnet = m*a So... m*a*ds = C * v^2 Nope...I really don't understand this, but it looks a little bit like the earlier equation. confidence assessment: 1
.................................................
......!!!!!!!!...................................
09:18:22 ** 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. **
......!!!!!!!!...................................
RESPONSE --> The very last bit loses me. Why does it follow that Fnet*ds = k*change in v^2 ? self critique assessment: 1
.................................................
......!!!!!!!!...................................
09:29:59 How do our experimental results confirm or refute this hypothesis?
......!!!!!!!!...................................
RESPONSE --> Didn't do an experiment vf^2 = v0^2 + 2a*ds vf^2-v0^2 = 2 a*ds If v0 ^2 = 0 then the change in velocity is = vf^2 so we can simplify and say vf^2 = 2 a * ds vf^2 / 2 = a*ds If we do an experiment where the slope is consant, but the distance is changed then the velocity should change in proportion to to a*ds. To confirm this we would graph vf^2 vs a*ds ( Even though I see the equation I am having a hard time really grasping why vf has to be squared for this). Since a is constant we can leave that out of the graph and just graph vf^2 vs ds and we should have a linear graph confidence assessment: 1
.................................................
......!!!!!!!!...................................
09:30:07 ** 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. **
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 0
.................................................