Work done by normal force on a block placed on movable inclined wedge A block placed on movable smooth inclined wedge placed on a smooth surface. Both are released and allowed to move.
I was told that

If the center of mass of wedge and block system is fixed (in the horizontal direction I assume, otherwise there won't be any motion) then in the journey of the wedge from A to B the Normal does not do any work because the displacement is point-wise perpendicular.

I do not understand Why is this so? Even if center of mass of the wedge and block is fixed the displacement of block $\vec{d}$ (as in the figure below) is not zero.
More formally if $d\vec{s}$ is
small displacement of the block we can write it as a sum of small displacement of the wedge $d\vec{s_1}$ and small displacement of the block $d\vec{r}$ wrt wedge.
$$\int_{A}^{B}\vec{N}.d\vec{s}=\int_{A}^{B}\vec{N}.d\vec{r}+\int_{A}^{B}\vec{N}.d\vec{s_1}$$
The first term is zero, but the second is not. Hence the work done by Normal must not be zero.
Am I misintrepretating something? Or what I was told was not correct? Any help will be highly appreciated.

This problem is quite similar. I read it's answer, but that still leaves the question unasnwered. The only thing it says about the work done by normal is

The vertical component of the force on the block due to the wedge N does negative work on the block but the horizontal component of force N does positive work on the block.

 A: The normal force indeed does work on the block.However, it does zero work on the block + wedge system:
Suppose the block gets displaced by $\vec{dr}$ relative to the wedge, and the wedge gets displaced by $\vec{dx}$. Then the block, as seen from the ground , is displaced by
$\vec{dr}$+$\vec{dx}$. The work done on the block=$dw1=\vec{N}.(\vec{dr}+\vec{dx})$, and the work done on the wedge = $dw2=(\vec{-N}).\vec{dx}$.
So after adding, we get $dw1+dw2$ = $\vec{N}.(\vec{dr})$=$\vec{0}$.
A: You are completely right in that the normal force will do work on the block. This is required so that the block and the wedge exchange energy. Your reasoning is the correct one, the motion of the block is not perpendicular to the normal force. This would only be the case in the frame of the wedge, which will in general not be inertial if the wedge is allowed to move.
With respect to the quote you mentioned, indeed, the normal force does not do any net work on the composite block-wedge system. This is essentially due to Newton's third law. It is for this reason that the total energy of the system is conserved, even though the normal force does non-conservative work on the block or the wedge, individually.
