asst 8 query

course Phy 201

10-03-2007?y??????J??????assignment #008

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008. `query 8

Physics I

10-03-2007

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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

this would be from the experiment done in the Class Notes.

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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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This looks very good. Let me know if you have any questions. &#