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,

enter image description here

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,

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

Conservation of energy of center of mass :

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

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


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

  • $\begingroup$ Newton's third law. $\endgroup$
    – march
    Commented Jun 23, 2023 at 20:17
  • $\begingroup$ @march how would that imply workdone to be zero? $\endgroup$
    – Chesx
    Commented Jun 23, 2023 at 20:20
  • 2
    $\begingroup$ Does this answer your question? Work done by normal force on a block placed on movable inclined wedge $\endgroup$
    – Amit
    Commented Jun 23, 2023 at 21:04
  • $\begingroup$ @Amit No it doesn't answer. $\endgroup$
    – Chesx
    Commented Jun 24, 2023 at 6:38
  • 1
    $\begingroup$ @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. $\endgroup$
    – Amit
    Commented Jun 24, 2023 at 11:19

1 Answer 1


Because both bodies move, the point of contact between them travels along a given path for an inertial observer at rest with the system before the movement starts.

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 $-N$.

So, the total work due to the Normal force is zero.

  • $\begingroup$ Who does work on behalf of normal? Which particle? $\endgroup$
    – Chesx
    Commented Jun 24, 2023 at 7:06
  • $\begingroup$ How could you say that path is same? Since there is not a fixed particle of wedge which move along the path. $\endgroup$
    – Chesx
    Commented Jun 24, 2023 at 7:34
  • $\begingroup$ The force does work while it moves. It is not necessary that the particle of matter in contact be the same. Think of a row doing work while cutting the water. $\endgroup$ Commented Jun 24, 2023 at 11:05

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