Is the normal force a conservative force?

Most of the time the normal force doesn't do any work because it's perpendicular to the direction of motion but if it does do work, would it be conservative or non-conservative?

For example, consider the following block-incline system where the incline is on a friction-less surface. Now as the block moves down, the incline itself starts moving in the opposite direction. Here, the normal force acting on the incline by the block does do work.

This question was on some website (it asked something about the final velocities of the block and the incline) and they solved this using mechanical energy and momentum conservation which confused me as I couldn't understand why the normal force here is conservative.

So,

1. Is the normal force always conservative or is this only in some cases?
2. If yes, how do we deduce that like in the above example?
3. Also, if it is conservative, what would be its corresponding potential energy function?
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BTW--a image would make it easier for people to understand your intent right away, which I think would get more attention because this is a good question. –  dmckee Jan 10 '13 at 16:10

The normal force acting on the incline by the block does do work, but the normal force acting on the block by the incline does negative work, and the total work done by all normal forces in the system is zero.

Edit - here's how we can see the work done by the two normal forces cancel (and this proof can be easily generalized to any other problem with normal forces):
According to Newton's second law, the force acting on the incline by the block ($\mathbf{N}_{bi}$) is equal in size and opposite in direction to the the force acting on the block by the incline ($\mathbf{N}_{ib}$), i.e.: $$\mathbf{N}_{bi} = - \mathbf{N}_{ib}.$$ Taking the horizontal component of both sides: $$N_{bi,x} = - N_{ib,x}.$$ The incline and the block move together horizontally, so if the incline travels an infinitesimal horizontal distance of $dx$, then the block at the same time travels the same distance $dx$. The total work done by both forces while this distance is traveled cancels: $$N_{bi,x} dx = - N_{ib,x} dx ~~~\Rightarrow~~~ N_{bi,x} dx + N_{ib,x} dx = 0$$

Therefore, the normal force can still be considered a "constraint force", i.e. a force that does no work and is neither conservative nor non-conservative.

What makes this example confusing is the fact that the work vanishes only when looking at all the normal forces in the system. This is because the normal force acts here as a mediating force, transferring the gravitational force from the block to the incline. A simpler example of such a force is the tension force of a string holding two weights over a pulley:

In this system the string pulls the lighter mass and does work on it, but it does negative work on the heavier mass and so the total work the tension forces do is zero. The string acts as a mediator that transfers the gravitational force between the two blocks.

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You make a good point but you haven't explained why the work done by the block on the inlcine is exactly equal to the work done by the incline on the block. I can see that the work is negative but I can't see if it's exactly equal. –  Alraxite Jan 15 '13 at 15:27
@Alraxite: Yes that explanation was missing from my answer, I have now added it. –  Joe Jan 15 '13 at 17:17

Depends on how you model it, and on the strain behavior of the bodies in question.

In the simplest models (it always, exactly and instantaneously balances the force holding the two bodeis together; i.e. the ususal physics 101 normal force) it is a "constraint force" and simply never does any work.

In more sophisticated models the bodies have non-trival strain moduli and the force can act over non-zero distances. In this case you would expect the force to be conservative (to first order it's just Hooke's law) while the bodies exhibit elastic behavior. Once you get into the non-elastic regime, thing are less clear. You can press two things together and on release not get all the work back (which sounds non-conservative) because that energy has gone into reconfiguring at least on of the bodies.

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So, is the normal force conservative and how do we know it? Also, my question doesn't regard bodies which can undergo strain. If it needs to be said, the objects are rigid. –  Alraxite Jan 10 '13 at 15:29
For rigid objects it is neither conservative nor non-conservative it is a "force of constraint" and never does any work. This concept comes up later in the context of Lagrangian mechanics, where it turns out to be very useful in making the need for understanding forces of constraint go away. –  dmckee Jan 10 '13 at 15:40
But clearly in my example, the normal force does do work. –  Alraxite Jan 10 '13 at 15:42
OK. I see where you are coming from. This is much clearer in the Lagrangian formulation. You've degomposed the weight into two bits $F_\perp$ and $F_\parallel$ just like in the case where the ramp doesn't more, but that's not really right, because the block isn't going to move down the angle of the ramp--the ramp is getting partially out of the way on the block goes more straight down. Moreover, because the ramp is pushed out from under the block the normal frce is reduced. –  dmckee Jan 10 '13 at 16:07
It should be possible to do a decomposition into a horizonal bit which pushes the ramp, a bit angled down along the boxes' trajectory and a bit canceled by the normal, but it is going to be a lot of work to get it exactly right. Or you could use Lagrangian mechanics, or you could say "there is no friction in the system, nor any discontinuous deformation, not losses to sound or heat so I expect conservative behavior." and go ahead. –  dmckee Jan 10 '13 at 16:09
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Normal force are reaction forces on solid things, like walls, and are perpendicular to this: If $dx$ has some perpendicular component to the surface, it must be outwards (it can't go through the surface), so the normal force dissapears as it stops touching the surface. If the normal always exists the object must move on the surface, then $dx$ is perpendicular to $N$ and the work made by that force is always 0. That means that: $$\oint_C Ndx=0$$ For every closed path $C$, as actually it will be 0 for every path, so it's conservative, but not very interesting as the work is always 0.

In moving reference systems, the normal is coupled to the force that keeps the object on the surface, so the work will be the same as the one by the force that keeps the object on the surface. It will be conservative if this force is.

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Though I appreciate you writing the answer, I really couldn't understand it! Could you please explain in a less mathematical way that how did you deduce the work done by the block on the incline is conservative here and hence we can use energy conservation to solve this kind of problem? –  Alraxite Jan 10 '13 at 15:44
Ok, I see. I would look at this problems not from the point of view of conservative forces, but conservation of energy. Energy is conserved always (universally), which means, the total energy before and after must be the same. If they tell you that there is no friction, then total energy of the system must be conserved, that is, initial potential energy of th block will be divided, part given to kinetic energy of the incline, and part given to kinetic energy of the block. That (friction) was the only way energy could "dissapear", and there's not friction, so energy must be conserved. –  MyUserIsThis Jan 10 '13 at 15:55
Btw, if you want a tool to know if a force is conservative, that's the rotational of the force: $\nabla\times F$. Even if you don't understand it yet (I guess you will study it in calculus), if the rotational of a force is 0, then the force is conservative. If it's different from 0, it's not. –  MyUserIsThis Jan 10 '13 at 15:57
Though the reasoning that there is no friction is fine, it's not a satisfactory way to show someone that the work done by the normal force here is conservative. Btw, I do know one way to show if a force is conservative: if we can assign a potential energy function to it. And I'm not sure how do we do that here. –  Alraxite Jan 11 '13 at 9:12
@Alraxite Yes, If you can assign a potential function $V$, then $F=-\nabla V$, and you can do that if and only if $\nabla\times F=0$, so if the curl is 0, then you can assign a potential and the force is conservative. The curl is a trivial calculation for most forces. Particularly, here that force is constant so the curl is 0 and it's conservative. –  MyUserIsThis Jan 11 '13 at 14:46