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I read somewhere that

work done by internal forces is independent of frame of reference

but suppose a man of mass $m$ is standing on a stationary smooth cart of mass $2m$, placed on a smooth surface. now assume the man jumps horizontally with relative velocity v with respect to cart.

in frame of cart, since no external work present, Work Done = $$\frac12 . m . v^2 $$ [note frame has no acceleration]

But by using momentum conservation we get the cart moves with velocity $-v/3$ and the velocity of man with respect to ground = $2v/3$

so using Work Energy Theorem in ground frame, work done by internal forces of man should be $$(\frac12 . m . \frac{4v^2}9) + (\frac12 .2m. \frac{v^2}9 ) = \frac{mv^2}3$$

Is my interpretation wrong? or is there any external force present that can have work?

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  • $\begingroup$ “In frame of block” Did you mean cart? $\endgroup$
    – Bob D
    Commented Jan 18, 2022 at 10:35
  • $\begingroup$ oh thanks, yes, it was cart. i have changed that $\endgroup$ Commented Jan 18, 2022 at 12:08

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The first frame of reference you use actually has to have acceleration since the man is jumping from the stationary cart they are accelerating. However because this requires them to exert a force on the cart and it does not have friction it should accelerate because of this force and move in the other direction, the frame then becoming the second frame. Hence your result in the first frame is incomplete

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