enter image description here

The given question is: A block of mass $m$ is placed on a triangular block of mass $M$ which is in turn placed on a horizontal surface (figure). Assuming frictionless surfaces, find the velocity of the triangular block when the small block reaches the bottom end.

I understand how to solve this in two ways:

  1. Finding the acceleration of the small block w.r.t the wedge, and using kinematics find the velocity of the smaller block once it completes a distance of $h/\sin α$. After that, we use the conservation of momentum along the horizontal axis to find the velocity of the bigger block.
  2. Alternatively, one can use conservation of momentum, then conserve mechanical energy. The potential energy of the smaller block will provide the kinetic energy of both the bigger block and the smaller block.

However, my interest is in solving this using the work-energy theorem.

My logic is as follows: the reaction force applied on the larger block ($R$) is equal to $mg\cos α$. The horizontal component of this force, $R\sin α$ is the only force acting in the horizontal direction. Then, apply the Work-Energy Theorem. We can find the displacement of the wedge easily (the horizontal component of the center of mass is conserved).

Then, multiplying the force and the displacement, we should get the change in kinetic energy. I'm not sure how to write the RHS (kinetic energy) of this equation. Thanks in advance.

The correct answer which agrees with methods 1 and 2 is:

$$u=\left[ \frac{2m^2 gh \cos ^2 \alpha}{(M+m)(M+m\sin^2 \alpha)}\right]^{\frac{1}{2}}$$

Please note that my question isn't on how to solve this problem generally, I've mentioned both the usual methods for solving a problem like this. My interest is in solving specifically it using the net horizontal force on the wedge multiplied by its displacement.

  • 3
    $\begingroup$ "the reaction force applied on the larger block (R) is equal to mgcos α" That isn't true here. $\endgroup$ Apr 7, 2020 at 20:31
  • $\begingroup$ I've removed a number of comments that were attempting to answer the question and/or responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. $\endgroup$
    – David Z
    Apr 8, 2020 at 9:53

2 Answers 2


The problem with your W.E.T. approach is that you assumed the reaction force ($R$) to be equal to $mg\cos \alpha$ which, in this case, is wrong. This assumption would have been true in this case:

enter image description here

Image source

Because in this case the motion is only along the inclined plane and there is no motion (and thus no acceleration) in the direction perpendicular to the plane. Which implies that all the forces along the perpendicular direction must cancel out.

But in your case, the incline plane is also moving. And thus from the ground frame, the block does have a non zero component of acceleration along the direction perpendicular to the inclined plane.

If you change your frame of reference to the non inertial frame of the inclined plane (although it won't help you solve this question using W.E.T. in any way because the displacement of the inclined plane will become zero in this frame), then in that frame, there would be no component of acceleration along the direction perpendicular to the inclined plane, thus in that frame you can balance the forces in the perpendicular direction. However, if you go to that non inertial frame of reference, you'd have to also apply a pseudo force on the block.


After some thinking and solving, I figured out the answer. Thanks FakeMod and Aaron for pointing out the flaw in my earlier logic. In short, the reaction force on the larger block isn't $mg\cos\alpha$, it'll change because of the acceleration on the system.

The solution I've come up with is as follows: A: acceleration of system leftwards N: reaction force between $m$ and $M$.

From the free bodies, we get: $$N=mg\cos\alpha- mA\sin\alpha$$ $$N\sin\alpha=MA$$

Solving these, we get: $$A= \frac {mg\cos\alpha\sinα}{M+m\sin^2α}$$

Now, since the horizontal coordinate of the COM is constant, we can write:

$$d = \frac{mh}{(M+m)\tanα}$$

We can now write the work energy theorem as follows:

$$(M+m)Ad= \frac 12(M+m)u^2$$

Substituting and solving, we get: $$u=\left[ \frac{2m^2 gh \cos ^2 \alpha}{(M+m)(M+m\sin^2 \alpha)}\right]^{\frac{1}{2}}$$ This agrees with the solution from other methods.

Despite this being a very simple problem, I haven't seen a similar approach elsewhere. Although it can be argued that the approach I've used reduces to the same equations as other methods, I feel such an approach is instructive in showing how many concepts can be used to solve this problem.

  • $\begingroup$ If you're anyways going to find the normal reaction, then you should directly do it using Newton's second law. That way is simpler. $\endgroup$
    – user258881
    Apr 8, 2020 at 6:07
  • $\begingroup$ Could you explain how to do that...? Also, thanks a lot for your answer, but I think we posted almost simultaneously... $\endgroup$
    – wavion
    Apr 8, 2020 at 6:09
  • $\begingroup$ How did you find the acceleration of the block in method 1? You'd have used Newton's second law, I suppose. And that's what I think is better if you are anyways going to write the force equations. But for exploratory purposes, this is a good method to explore and learn. $\endgroup$
    – user258881
    Apr 8, 2020 at 6:11
  • 1
    $\begingroup$ Oh right, of course. My approach is a little convoluted, but it came up naturally when I was writing equations in the wedge frame. Thanks a lot for your help, cheers! $\endgroup$
    – wavion
    Apr 8, 2020 at 6:15

Not the answer you're looking for? Browse other questions tagged or ask your own question.