It is evident from these queries and responses that the Work Energy Theorem holds true in non-inertial frames (with work performed by pseudo force included in the equation).

  1. Is work-energy theorem valid in non-inertial frames?
  2. Work associated with pseudo force

Now if we Consider the following situation of a closed system :

enter image description here

As shown, a block (Objectโ€“A) of mass mA=m moves on a frictionless plane at initial speed v0, and lands onto a cart (Objectโ€“B in the figure, mass mB=m, length L, initially at rest) smoothly. Ignore the size of the block. The friction coefficient between A and B is u and the cart is on a frictionless plane. The block collides elastically on the fixed wall w.r.t. cart at the end of the cart and eventually falls off the cart. Suppose the initial speed is uo then loss/change in the kinetic energy of the system ( block and Cart) is ? ( no external forces)

( Note: Now, this may also be solved using ground frame by figuring out the net work produced by kinetic friction (only working force) and relative displacement.)

If we utilize the Work-Energy Theorem from the cart's (non-inertial of course) frame of reference to solve the problem, a Pseudoforce equal in magnitude to the frictional force acts on the block in the direction counter to vo.

Now During the whole motion, only 2 forces acts on the block, pseudo force and kinetic friction :

  1. work done by frictional force acting on block due to cart : ๐‘Š๐‘“= -2๐‘š๐‘ข๐‘”๐‘™
  2. work done by pseudo force on the block : ๐‘Šโ‚š = -2๐‘š๐‘ข๐‘”๐‘™
  3. No work is being done on the cart in this frame ( No net forces )

![enter image description here

Now, using work energy theorm, Change in kinetic energy of the system must be equal to the total work done by all forces. Therefore,

W\index{p} +W\index{\index{f} }= change in kinetic enegery = -4mugl

Furthermore, if we use the laws of motion and the conservation of linear momentum to calculate the starting and end velocities (relative to the cart), the change in kinetic energy relative to the cart is in fact -4mugl from this frame.

However, this is incorrect. The right solution to the problem was -

enter image description here

Which does makes sense to me, but my doubts are

  1. What's wrong with utilizing WET in a non-inertial frame? What Am I missing ?

  1. Is the calculated change in system kinetic energy (-4mugl) not the real change in KE? So what exactly is it?

Any response that aims to enlighten me on these matters is greatly appreciated.

  • $\begingroup$ Why did you take friction = $2mgu$? $\endgroup$
    – khaxan
    Oct 1, 2023 at 16:10
  • $\begingroup$ @khaxan the value of friction is mgu, Actually the total length travelled is "2L" $\endgroup$
    – TPL
    Oct 1, 2023 at 16:23
  • $\begingroup$ How is the problem formulated? What loss in KE? Of what? $\endgroup$
    – nasu
    Oct 1, 2023 at 19:10
  • $\begingroup$ @nasu Loss of Kinetic Energy of the Block and Cart System $\endgroup$
    – TPL
    Oct 2, 2023 at 4:36

1 Answer 1


You can solve the problem by working in a non-inertial frame, but it is important to later calculate the velocities in the "ground" frame to determine the change in kinetic energy in this frame. The issue is that while the work-energy theorem works fine in non-inertial frames, the work done (i.e. the change in energy) is not a frame-independent quantity.

Suppose you hit a tennis ball initially at rest, imparting to it a kinetic energy of $E$ in the (inertial) rest frame. What happens in the non-inertial frame moving with the tennis ball? Here, the force of you hitting the ball and the pseudo-force cancel out exactly, so no work is done. This is consistent with the fact that the ball remains at rest in this frame, but the corresponding change in kinetic energy (i.e. 0) is not the same as that in the inertial frame.


Your Answer

By clicking โ€œPost Your Answerโ€, you agree to our terms of service and acknowledge you have read our privacy policy.

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