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Question: A block of Mass m is connected to another block of mass M by a mass-less spring of spring constant k . The blocks are kept on a smooth horizontal plane and are at rest. The spring is unstretched when a constant force F starts acting on the block of mass M(horizontally) to pull it. Find the maximum extension of the spring.

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Here I assumed that max. extension(x) will be produced when both the blocks would be moving with constant and same acceleration. Then by considering the FBD diagram of the blocks m and M, I formed this equation

$\Large \frac{kx}{m}= \frac{F-kx}{M}$

and hence I get the value of x as

$\Large x=\frac{mF}{k(m+M)}$

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by conservation of energy In the reference frame of center of mass.

For block 'm', we have two forces acting $mF/(m+M)$ and $kx$, in opposite directions.

For block 'M' we have three forces acting $MF/(m+M)$ and $kx$, and $F$ in the opposite direction.

Assuming m moves a max distance x_1 from CM M moves a distance x_2 from CM. Then work done by external force will be

$$W=mF(x_1+x_2)/(m+M)$$

This will be stored as the internal energy, therefore

$$1/2·k·(x_1+x_2)^2=W$$

On solving this we get

$$x_1+x_2=2mF/(k(m+M))$$

But my method is not correct. Can someone please help me out understand why this approach is not correct?

(When I looked up for the solution, the correct answer given was $\large \frac{2mF}{k(m+M)}$, by solving it from reference frame as COM, by using work-energy theorem but if that is the answer both the blocks would have different accelerations .)

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  • $\begingroup$ I think there is a typo in "Assuming m moves a max distance x_1 from CM M moves a distance x_2 from CM." $\endgroup$ – DKNguyen Nov 18 at 15:12