# Why work done by internal force is changing with frame in this case?

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?

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