friction

[thebillsfan]

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I got these from EK:

quote: W=Fdcos(theta) is good for all forces except friction

Why is friction an exception here? Let's say d represented distance and not displacement--why couldnt we use this equation to calculate the work done by friction?

quote: The work done by friction can only be found if the change in internal energy is known

Again, why? If you know the force of friction and the distance the object traveled, can you not calculate the work done by fric?

Finally:

quote: Except for frictional forces, the work done by all nonconservative forces equals the change in the mechanical energy of the systems upon which they are applied.

Why isn't the work done by fric also equal to the change in the mechanical energy? It should be like any other noncons. force.


I took these EK notes a long time ago, when I may have interpreted the book wrong. But I'm pretty sure that's what the book was saying.

thanks in advance!

 

[wanderer]

I got these from EK:

quote: W=Fdcos(theta) is good for all forces except friction

Why is friction an exception here? Let's say d represented distance and not displacement--why couldnt we use this equation to calculate the work done by friction?

quote: The work done by friction can only be found if the change in internal energy is known

Again, why? If you know the force of friction and the distance the object traveled, can you not calculate the work done by fric?

Finally:

quote: Except for frictional forces, the work done by all nonconservative forces equals the change in the mechanical energy of the systems upon which they are applied.

Why isn't the work done by fric also equal to the change in the mechanical energy? It should be like any other noncons. force.


I took these EK notes a long time ago, when I may have interpreted the book wrong. But I'm pretty sure that's what the book was saying.

thanks in advance!

This is all wrong. W=Fdcos(theta) for all forces. Friction converts mechanical energy (PE and KE) into thermal energy effectively "wasting" it. If you still have your exams from physics 1 there's probably some question about, "Give the work due to friction," as it is a common physics one question.

 

[thebillsfan]

alright. but what about the scenario where work done by friction not only decreases mechanical energy but increases internal energy? if youre not given the change in internal energy, how can you figure out what portion of the friction work went to mech. energy and what portion went to internal?

 

[Omni]

You can use that equation, so long as d=distance and not displacement. If you move a box on a carpet 10 meters north, and then 10 meters south, the displacement is 0, but that doesn't mean the work done by friction turns into 0 as well. However, distance is 20, so it would F*(cosX)*20, or just F*20, since frictional always opposes motion and the box is being moved parallel to the surface.

 

[supras2kracer]

Sorry to open this back up but I have the same exact question as the OP.

In my physics class we learned that the work from friction is just Fd. Most physics forums I found online all say the same thing. What the heck is EK talking about with this internal energy stuff?

 

[Akarat]

Sorry to open this back up but I have the same exact question as the OP.

In my physics class we learned that the work from friction is just Fd. Most physics forums I found online all say the same thing. What the heck is EK talking about with this internal energy stuff?

I'm going to say that this has something to do with the fact that friction produces heat, and that a system that experiences friction will increase in internal energy because of this heat.

Based on the First Law of Thermodynamics:
[delta]E = q - w
where [delta]E is the change in the internal energy of the system, q is heat, and w is work.

Since heat is being transferred from the surroundings to the system, q is positive, and thus [delta]E increases.

Anyone know whether w is positive or negative in this case?

 

[Isoprop]

The EK book is wrong or going beyond the scope of the MCAT.

You can absolutely calculate the work due to friction with the same equation: Fdcosθ. cosθ is always -1 because the direction of motion is always the opposite of the force of friction (θ=180 degrees).

Thus, kinetic friction is always doing negative work. What does negative work mean? It means that work is done to decrease the mechanical energy of the system. This is why a hockey puck will lose kinetic energy because of friction. The friction is doing negative work and decreasing the overall kinetic energy.

Remember, friction doesn't always produce energy in the form of heat-- it can also produce energy in the form of sound, light, etc.

Also remember that static friction does no work because the displacement/distance is always zero.

 

[supras2kracer]

The EK book is wrong or going beyond the scope of the MCAT.

You can absolutely calculate the work due to friction with the same equation: Fdcosθ. cosθ is always -1 because the direction of motion is always the opposite of the force of friction (θ=180 degrees).

Thus, kinetic friction is always doing negative work. What does negative work mean? It means that work is done to decrease the mechanical energy of the system. This is why a hockey puck will lose kinetic energy because of friction. The friction is doing negative work and decreasing the overall kinetic energy.

Remember, friction doesn't always produce energy in the form of heat-- it can also produce energy in the form of sound, light, etc.

Also remember that static friction does no work because the displacement/distance is always zero.

Ok, thanks for clearing that up. I think exam krackers is wrong. The audio osmosis guys said the same thing as the book and it made even less sense.