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.


  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?
  • 2
    $\begingroup$ 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. $\endgroup$ Jan 10, 2013 at 16:10
  • 3
    $\begingroup$ The concept of conservative versus nonconservative forces is only of interest when the force is a field of force, i.e., a force whose value depends only on the position of the object being acted on. $\endgroup$
    – user4552
    Oct 3, 2014 at 0:55

2 Answers 2


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: see below for proof).
Therefore, the normal force can be considered a "constraint force", i.e. a force that does no work and is neither conservative nor non-conservative.
The work vanishes only when looking at all the normal forces in the system, since the normal force acts here as a mediating force, transferring the gravitational force from the block to the incline.

This example may be confusing since there are additional forces in different directions, consider the simpler setting of a force pushing two blocks on a horizontal plane:

Force pushing two blocks on a horizontal plane
Here the left block applies a normal force to the right block and vice versa, and again the total work done by the two normal forces cancels, since the normal force mediates the pushing force between the left block and the right block.

Another interesting example is the tension force of a string holding two weights over a pulley:

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.

Edit - corrected proof (credit to @DSinghvi for pointing out the error in the previous version of the proof in the comment below):
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 force acting on the block by the incline, $\mathbf{N}_{ib}$, i.e.: $$ \mathbf{N}_{bi} = - \mathbf{N}_{ib}. $$ In the axis parallel to the normal force, the incline and the block move together, so if the incline travels an infinitesimal 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: $$ N_{bi} dx = - N_{ib} dx ~~~\Rightarrow~~~ N_{bi} dx + N_{ib} dx = 0 $$

  • $\begingroup$ 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. $\endgroup$
    – Alraxite
    Jan 15, 2013 at 15:27
  • $\begingroup$ @Alraxite: Yes that explanation was missing from my answer, I have now added it. $\endgroup$
    – Joe
    Jan 15, 2013 at 17:17
  • 2
    $\begingroup$ @Joe How are the displacement going to be same. I mean they will be governed by the ratio of their masses and the state of motion of Centre of mass. Even though if I assume masses are same wouldn't it be that the normal force are acting in the direction of moton of two bodies . One is moving backwards and other forward. $\endgroup$
    – DSinghvi
    Dec 29, 2014 at 9:53
  • $\begingroup$ Wow, indeed there was an error in my proof. I can't believe it was there so long before someone noticed it. @DSinghvi - thanks for pointing it out and good catch! I corrected the error and also improved the answer a bit. $\endgroup$
    – Joe
    Dec 29, 2014 at 19:37

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.

  • $\begingroup$ 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? $\endgroup$
    – Alraxite
    Jan 10, 2013 at 15:44
  • $\begingroup$ 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. $\endgroup$ Jan 10, 2013 at 15:55
  • 1
    $\begingroup$ 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. $\endgroup$ Jan 10, 2013 at 15:57
  • $\begingroup$ 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. $\endgroup$
    – Alraxite
    Jan 11, 2013 at 9:12
  • 1
    $\begingroup$ @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. $\endgroup$ Jan 11, 2013 at 14:46

Not the answer you're looking for? Browse other questions tagged or ask your own question.