asst 8 query

course Phy 121

assignment #008008. `query 8

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

Physics I

06-14-2007

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10:52:56

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

See if you can answer this question.

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RESPONSE -->

Force is any kind of push or pull on an object. Fnet is the sum of the forces acting on an object.

confidence assessment: 3

.................................................

......!!!!!!!!...................................

10:53:09

** 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 -->

ok.

self critique assessment: 3

.................................................

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10:56:35

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 -->

I know that you can get average change in velocity from acceleration and change in distance. But I am not sure why v^2 is proportional to Fnet and `ds.

confidence assessment: 2

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11:03:08

** 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 -->

I think I am starting to understand these concepts. I am not 100% sure about them. Is there another way to explain this so that maybe I can understand it a little more?

self critique assessment: 2

A briefer explanation:

vf^2 = v0^2 + 2 a `ds (equation of motion) and

a = F_net / m (Newton's Second Law) so

vf^2 = v0^2 + 2 F_net / m * `ds, which can be solved for F_net `ds to get

F_net * `ds = 1/2 m ( vf^2 - v0^2).

Thus F_net * `ds is proportional to vf^2 - v0^2, which is the change in the squared velocity.

The proportionality constant is 1/2 m.

.................................................

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11:04:01

How do our experimental results confirm or refute this hypothesis?

......!!!!!!!!...................................

RESPONSE -->

I am not sure what experimental results we are talking about and not sure what the hypothesis is either.

confidence assessment: 0

.................................................

......!!!!!!!!...................................

11:06:12

** 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 -->

I get if a`ds is proportional to the change in vf^2 then the graph should be linear.

self critique assessment: 3

.................................................

"

&#

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

asst 8 query

course Phy 121

assignment #008008. `query 8

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

Physics I

06-14-2007

......!!!!!!!!...................................

10:52:56

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

See if you can answer this question.

......!!!!!!!!...................................

RESPONSE -->

Force is any kind of push or pull on an object. Fnet is the sum of the forces acting on an object.

confidence assessment: 3

.................................................

......!!!!!!!!...................................

10:53:09

** 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 -->

ok.

self critique assessment: 3

.................................................

......!!!!!!!!...................................

10:56:35

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 -->

I know that you can get average change in velocity from acceleration and change in distance. But I am not sure why v^2 is proportional to Fnet and `ds.

confidence assessment: 2

.................................................

......!!!!!!!!...................................

11:03:08

** 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 -->

I think I am starting to understand these concepts. I am not 100% sure about them. Is there another way to explain this so that maybe I can understand it a little more?

self critique assessment: 2

A briefer explanation:

vf^2 = v0^2 + 2 a `ds (equation of motion) and

a = F_net / m (Newton's Second Law) so

vf^2 = v0^2 + 2 F_net / m * `ds, which can be solved for F_net `ds to get

F_net * `ds = 1/2 m ( vf^2 - v0^2).

Thus F_net * `ds is proportional to vf^2 - v0^2, which is the change in the squared velocity.

The proportionality constant is 1/2 m.

.................................................

......!!!!!!!!...................................

11:04:01

How do our experimental results confirm or refute this hypothesis?

......!!!!!!!!...................................

RESPONSE -->

I am not sure what experimental results we are talking about and not sure what the hypothesis is either.

confidence assessment: 0

.................................................

......!!!!!!!!...................................

11:06:12

** 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 -->

I get if a`ds is proportional to the change in vf^2 then the graph should be linear.

self critique assessment: 3

.................................................

"

&#

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