Question

A block of mass $m$ is connected to another block of mass $M$ by a massless 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 the block of mass $M$ to pull it. Find the maximum extension of the spring.

Answer 1

We solve the situation in the reference frame of centre of mass. As only $F$ is the external force acting on the system, due to this force, the acceleration of the centre mass is $Fl(M+m)$. Thus, with respect to centre of mass, there is a pseudoforce on the two masses in the opposite direction, the free body diagram of $m$ and $M$ with respect to centre of mass (taking centre of mass at rest) in showing Fig.
Taking centre of mass at rest, if $m$ moves maximum by a distance $x_1$ and $M$ moves maximum by a distance $x_2$, then the work done by external forces (including pseudoforce) will by
$W=\frac{mF}{m+M}{x}_1+(F+\frac{MF}{m+M})x_2=\frac{mF}{m+M}(x_1+x_2)$
This work is stoned in the form of potential energy of the spring as $U=\left(1/2\right)k(x_1+x_2{)}^2$.
Thus, on equating we get the maximum extension in the spring, as after this instant the spring starts contracting.
$\frac{1}{2}k(x_1+x_2{)}^2=\frac{mF}{m+M}(x_1+x_2)$
$⟹x_{max}=x_1+x_2=\frac{2mF}{k(m+M)}$