Is the normal force a conservative force?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.