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Using simple relative motion and Newton's laws of motion, I could prove that the work done by the normal force on a (unmounted straight inclined wedge + block) as a whole system is zero. Similarly, can it be proven mathematically that as the block slides down an unmounted curved inclined plane, the work done by the normal force on the whole system from the ground frame is zero? (The wedge is free to move on the horizontal surface and all surfaces are frictionless.)

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  • $\begingroup$ Which normal force are you asking about? $\endgroup$
    – Dale
    Jul 26 at 3:20
  • $\begingroup$ @Dale The normal force between the two bodies, acting during the block moves down the wedge. $\endgroup$
    – Ayush Naman
    Jul 26 at 3:48
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    $\begingroup$ You don't have to prove anything. The normal force is always perpendicular to the surfaces in contact, here, and the direction of motion is always parallel to the surfaces in contact, so the normal force does no work in this situation. (Well, provided that the wedge doesn't move.) $\endgroup$
    – march
    Jul 26 at 5:04
  • $\begingroup$ The curved inclined surface can be seen as the limit of a broken line. There're no essential differences. for mathematical rigor, all you need to do is figuring out the slope at each point on the curve. $\endgroup$
    – Luessiaw
    Jul 26 at 5:44
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    $\begingroup$ @BowlOfRed Think of a bomb (the system), with no external forces acting, exploding. $\endgroup$
    – Farcher
    Jul 26 at 9:40

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Similarly, can it be proven mathematically that as the block slides down an unmounted curved inclined plane, the work done by the normal force on the whole system from the ground frame is zero?

Yes offcourse, now I am providing you a brief solution.

enter image description here enter image description here enter image description here

If you want to find out complete work by normal on m and M , I am providing you expression for normal force , find work by integration. enter image description here

I hope it's clear to you, if not ask in comment.

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