Question

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

• A. $\frac{2mF}{3k(m+M)}$
• B. $\frac{mF}{k(m+M)}$
• C. $\frac{2mF}{k(m+M)}$
• D. $\frac{mF}{2k(m+M)}$

Answer 1

The correct option is C $\frac{2mF}{k(m+M)}$

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 of mass is $\frac{F}{M+m}$. Thus with respect to centre of mass, there is as pseudo force on the two mases in the opposite direction, the free body digram of m and M with respect to centre of mass (taking centre of mass at rest ) is shown in figure.

Taking centre of mass ast 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 be
$W=\frac{mF}{m+M}{x}_1+(F-\frac{MF}{m+M})x_2=\frac{mF}{m+M}(x_1+x_2)$ this work si stored in the form of potential energy of the spring as
$U=\left(\frac{1}{2}\right)k(x_1+x_2{)}^2.$
Thus, on equation we get the maximum extension in the spring, as after this instant the spring starts contracting.
$\frac{1}{2}k(x_1+x_2{)}^2=\frac{mF}{m+M}(x_1+x_2)$
$\Rightarrow x_{max}=x_1+x_2=\frac{2mF}{k(m+M)}$