Normal force, work and conservativity

I have searched very much on line, both in this site and elsewhere, but found no proof of whether the normal force is conservative or is not, in general.

Clearly, if the force is orthogonal to the velocity, $\mathbf{F}\cdot d\mathbf{r}=\mathbf{F}\cdot\mathbf{v}dt=0$, and the force does no work, as explained for example here, so it is conservative in that sense in that case.

But what happens if the normal force does work, as it can happen to do? And how can what happens be mathematically proved? I heartily thank you for any answer!

• It would be interesting to know the reason of the downvote: have I stated anything incorrect? Thank you very much! Apr 17 '15 at 17:06
• down votes rarely get a comment. don't worry about it Apr 20 '15 at 19:42

$\vec F \cdot \mathrm{d} \vec r$ does not determine if a force is conservative or not. All normal forces (conservative or not) produce work equal to $\vec F \cdot \mathrm{d} \vec r$ but what determines if they are conservative is the integral of $\vec F \cdot \mathrm{d} \vec r$ in a closed loop. If that is equal to zero i.e. if $\oint \vec F \cdot \mathrm{d} \vec r = 0$ then it is conservative because no energy is lost in that loop. It is like gravity. You throw something upwards and no energy is lost(in a vacuum). So the integral I mentioned is zero.