# What is the reason of work done by normal force zero? [closed]

Recently I was solving a problem involving a block on triangular wedge kept on horizontal fricitionless surface and the system is released from rest. All things are indicated in figure,

Now I am finding final velocities of blocks after block reaches ground using several methods.

$$N$$ represents normal between wedge and block and $$N_s$$ represents the normal between wedge and surface, $$x_1$$ represents horizontal displacement of $$m$$, $$x_2$$ represents horizontal displacement of $$M$$, $$v_1$$ & $$v_2$$ are final velocities of $$m$$ & $$M$$ respectively.

Appling work energy theorem :

\begin{align} N\cos\alpha\cdot x_1+(mg-N\sin\alpha)\cdot h+N\cos\alpha\cdot x_2+(N_s-N\sin\alpha)\cdot 0+Mg\cdot 0= \frac{mv_1^2}{2}+ \frac{Mv_2^2}{2} -0 \end{align}

Rearranging this gives,

$$$$\tag{1} mgh+N((x_1+x_2)\cos\alpha-h\sin\alpha)=\frac{mv_1^2}{2}+ \frac{Mv_2^2}{2}$$$$

Conservation of energy of center of mass :

$$\tag{2} mgh=\frac{mv_1^2}{2}+ \frac{Mv_2^2}{2}$$

By $$(1)$$ and $$(2)$$

$$N((x_1+x_2)\cos\alpha-h\sin\alpha)=0$$

So, this gives me that work done by normal is zero. But this was mathematics not physics.

What is exactly workdone by normal force?

Since point of contact changes every time as the block goes down, simply there is not a single particle which exerts normal on the other particle of block.

And can you tell me why this thing comes to be zero?

In the other words,

Why is the workdone by normal on wedge is of opposite sign and equal in magnitude to that of workdone by normal on block?

Please try to use mathematics as less as possible

• Newton's third law. Commented Jun 23, 2023 at 20:17
• @march how would that imply workdone to be zero? Commented Jun 23, 2023 at 20:20
• Does this answer your question? Work done by normal force on a block placed on movable inclined wedge
– Amit
Commented Jun 23, 2023 at 21:04
• @Amit No it doesn't answer. Commented Jun 24, 2023 at 6:38
• @Chesx BTW, I'm not sure but I think you may be decomposing the normal force wrongly along the $x$ and $y$ axes. Namely, I think you have your sines and cosines opposite to what they should be. Please see this diagram to understand what I mean.
– Amit
Commented Jun 24, 2023 at 11:19

The work done on the block due to $$N$$ is the integral of the dot product between $$N$$ and the block displacement along the contact path.
But the work done by the block on the wedge is the same integral because the path is the same, except for the force, that instead of $$N$$ is $$-N$$.