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

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

$\large Solution$

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


This will be stored as the internal energy, therefore


On solving this we get


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 .)

  • $\begingroup$ 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