q14 VCAA 2011

[VCE_2012]

This question was easy.
Suppose there WAS a constant friction force of X newtons acting in the opposite direction of the sled. What would be the momentum then?

Correct me if I'm wrong:
You start off by=>
mgh-fr*h=0.5mv^2 (not h, but the distance it traveled)
fr= force of friction.

[StumbleBum]

The loss of energy due to F_r wouldn't be F_r*h, it would F_r*The distance from AB

[VCE_2012]

The loss of energy due to F_r wouldn't be F_r*h, it would F_r*The distance from AB

There maybe a question like this, this year.

[StumbleBum]

I would assume if they were going to put in a question that is similar to that, they would make it a question that is more force related where you would need to find F_d(Force down the plane) and it would involve F_r and you would have to find an acceleration or something.

There maybe a question like this, this year.

Oh yeh, and what makes you think this, just out of curiosity?

[VCE_2012]

I would assume if they were going to put in a question that is similar to that, they would make it a question that is more force related where you would need to find F_d(Force down the plane) and it would involve F_r and you would have to find an acceleration or something.

There maybe a question like this, this year.

Oh yeh, and what makes you think this, just out of curiosity?

I was reading the examiner's report and realized that this question was handled pretty well, since it was handled pretty well there's a chance that they will hype  it up to tests people's knowledge in the physics involved. Well, that's what I believe.

[VCE_2012]

I would assume if they were going to put in a question that is similar to that, they would make it a question that is more force related where you would need to find F_d(Force down the plane) and it would involve F_r and you would have to find an acceleration or something.

There maybe a question like this, this year.

Oh yeh, and what makes you think this, just out of curiosity?

In the scenario you described the 'mgh' would still be considered right?

[Aurelian]

This question was easy.
Suppose there WAS a constant friction force of X newtons acting in the opposite direction of the sled. What would be the momentum then?

Correct me if I'm wrong:
You start off by=>
mgh-fr*h=0.5mv^2
fr= force of friction.

Friction acts always opposite (and therefore parallel) to the direction of motion. Thus the work done by the frictional force will be F * d, where d is the distance you actually cover (i.e. on the slope), not the height. This is contrasted with your change in potential energy; U depends only on where you are now - i.e. your current position - whereas the work done by friction depends on the path you've taken.


EDIT: Lol, 5 other people totally replied between my reply... Oh well I'll post anyway =3

[StumbleBum]

No, because im suggesting it would be a question where we would have to find the acceleration of say a box going down an inclined slope with F_r=xxxN so you would need to resolve the weight force into its respected component down the plane and then find the F_net to be able to solve for acceleration.
They could also include something to do with tension when something is towing something else to increase difficulty, which they would most likely do.

There is a question similar to what im talking about in the NEAP 2012 practice exam, although it is constant speed so theres no acceleration.

[VCE_2012]

No, because im suggesting it would be a question where we would have to find the acceleration of say a box going down an inclined slope with F_r=xxxN so you would need to resolve the weight force into its respected component down the plane and then find the F_net to be able to solve for acceleration.
They could also include something to do with tension when something is towing something else to increase difficulty, which they would most likely do.

It would be funny if they added a spring 3 meters (of constant friction) from the base of the slope or a cliff at the top of the incline XO