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I've been given two conceptual problems that I can't quite crack, at least, I'm not sure that I'm cracking them:
Problem 1
When a box is moving across a rough surface the friction force does negative work. Can a friction force ever do positive work? If yes, give an example, if no, explain why.
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I've aserted, in my so-far solution, that friction force can never do positive work, based on rudimentary readings on the perpetuum mobile (the eternity machine). The main reason for it's unattainability is the existence of friction, thus I don't think friction can do positive work. Since it would result in an increase in kinetic energy, without having anything to transform it from. I am, however, not sure that the arguments hold. So I'm now chasing either the right arguments or the one example that will prove friction force can do positive work.
If a box is on a conveyor belt, and the conveyor belt is turned on, and thus accelerates, the box stays for a while (because of static friction), but is accelerating along with the belt. At some point the box starts sliding across the belt, and there is kinetic friction. But - since the absence of kinetic friction would cause the box to stay in it's initial position (with respect to the inertial coordinate system - the Earth), isn't the existence of kinetic friction then directed like the movement itself, thus doing positive work on the box?
Problem 2
A rope is used to pull a box, which therefore accelerates. According to Newtons 3rd Law the box pulls the rope with a force as big but with opposite direction. Is the total work done, as mentioned in the work-energy theorem, then 0? If it is, how can you change the kinetic energy of a body?
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The work-energy theorem says that W.tot=K2-K1=deltaK. But this applies to the work of one body on another body, not to the total work of a system. If I pull the box, I do the work W.tot, which it counters with -W.tot (the same size force, with the opposite direction). The total value of work will then, always, be 0, since W.tot+(-W.tot) = 0. But isn't that addition quite irrelevant? The kinetic energy of both bodies has been altered, which can easily by seen by using the work-energy theorem.
Is this some kind of trick question?
Any help would be muchly appreciated. The assignment's due tomorrow.
Problem 1
When a box is moving across a rough surface the friction force does negative work. Can a friction force ever do positive work? If yes, give an example, if no, explain why.
-----
I've aserted, in my so-far solution, that friction force can never do positive work, based on rudimentary readings on the perpetuum mobile (the eternity machine). The main reason for it's unattainability is the existence of friction, thus I don't think friction can do positive work. Since it would result in an increase in kinetic energy, without having anything to transform it from. I am, however, not sure that the arguments hold. So I'm now chasing either the right arguments or the one example that will prove friction force can do positive work.
If a box is on a conveyor belt, and the conveyor belt is turned on, and thus accelerates, the box stays for a while (because of static friction), but is accelerating along with the belt. At some point the box starts sliding across the belt, and there is kinetic friction. But - since the absence of kinetic friction would cause the box to stay in it's initial position (with respect to the inertial coordinate system - the Earth), isn't the existence of kinetic friction then directed like the movement itself, thus doing positive work on the box?
Problem 2
A rope is used to pull a box, which therefore accelerates. According to Newtons 3rd Law the box pulls the rope with a force as big but with opposite direction. Is the total work done, as mentioned in the work-energy theorem, then 0? If it is, how can you change the kinetic energy of a body?
-----
The work-energy theorem says that W.tot=K2-K1=deltaK. But this applies to the work of one body on another body, not to the total work of a system. If I pull the box, I do the work W.tot, which it counters with -W.tot (the same size force, with the opposite direction). The total value of work will then, always, be 0, since W.tot+(-W.tot) = 0. But isn't that addition quite irrelevant? The kinetic energy of both bodies has been altered, which can easily by seen by using the work-energy theorem.
Is this some kind of trick question?
Any help would be muchly appreciated. The assignment's due tomorrow.
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