# 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.

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 $$dw1+dw2$$ = $$\vec{N}.(\vec{dr})$$=$$\vec{0}$$.