query 8 

course phy 201

???z??y{Y??????assignment #008

008. `query 8

Physics I

02-28-2009

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20:48:18

QUESTION FROM STUDENT--Please define the differnece between Fnet and Force.

See if you can answer this question.

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RESPONSE -->

Fnet = sum of all forces on an object

Force = any kind of push or pull on a object

confidence assessment: 3

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20:48:37

** 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 -->

self critique assessment:

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20:51:23

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 -->

F = ma

a = dv/dt

Becaue velocity and acceleration are directly related so a change in one will effect the other. Velocity does not change unless there is a force acting on it therefore F and V are related as well

confidence assessment: 1

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20:57:19

** 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 dont understand where we get the k? I understand its a new variable but I am used to sub. in only for F = ma so we are ultimatly sub. in k for mass?

self critique assessment: 2

k is the proportionality constant, which for a given situation can be anything at all so long as it doesn't vary.

The above shows that a `ds = k ( change in v^2) for k = 1/2.

From the fact that F = m a, it follows that Fnet `ds = k * (change in v^2) for k = mass / 2.

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21:04:30

How do our experimental results confirm or refute this hypothesis?

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RESPONSE -->

experiment 1 shows that mass of object and if appliable slope an object is traveling down both are important factors in det. Force. and acceleration

confidence assessment: 1

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21:09:36

** 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 -->

a will be constant 9.8m/s? if that is true then it should follow the hypothesis of an increase in slope would decrease time and velocity would increase because the distance would be convered in less time. I understand why vf would be the only variable since v0 would be recorded as 0.

I still dont see where we are getting a*ds and setting it equal to vf^2

self critique assessment: 2

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According to the fourth equation of motion vf^2 = v0^2 + 2 a `ds.

Thus if v0 = 0, as is the case here, vf^2 = 2 a `ds.

Thus a graph of vf^2 vs. `ds will have slope 2a.

&#Good work. See my notes and let me know if you have questions. &#