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. Initially, the blocks are at rest and the spring is un-stretched when a constant force F starts acting on the block of mass M to pull it. Find the maximum extension of the spring?

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

Answer 1

The correct option is C

$\frac{2mF}{k(m+M)}$

Let us take the two blocks plus the spring as the system. The centre of mass of the system moves with an acceleration $a=\frac{F}{m+M}$ Let us work from a reference frame with its origin at the centre of mass. As this frame is accelerated with respect to the ground we have to apply a pseudo force ma towards left on the block of mass m and Ma towards left on the block of mass M. The net external force on m is

$F_1=ma=\frac{mF}{m+M}$ towards left

and the net external force on M is

$F_2=F-Ma=F-\frac{MF}{m+M}=\frac{mF}{m+M}$ towards right.

The situation from this frame is shown in figure. As the centre of mass is at rest in this frame, the blocks move in opposite directions and come to instantaneous rest at some instant. The extension of the spring will be maximum at this instant. Suppose the left block is displaced through a distance $x_1$ and the right block through a distance $x_2$ from the initial positions. The total work done by the external forces $F_1$ and $F_2$ in this period are

$W=F_1{x}_1+F_2{x}_2=\frac{mF}{m+M}(x_1+x_2)$.

This should be equal to the increase in the potential energy of the spring as there is no change in the kinetic energy. Thus,

$\frac{mF}{m+M}(x_1+x_2)=\frac{1}{2}k(x_1+x_2{)}^2$

$x_1+x_2=\frac{2mF}{k(m+M)}$

This is the maximum extension of the spring.