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 of 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

Explanation

Let us take the two blocks plus the spring as the system. The centre of mass of the system moves with an acceleration `a=(F)/(m+M)`. Let us work from a reference frame with its origin at the centre of mass. As this frame is acceleration 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=(mF)/(m+M)` towards left
and the net external force on `M` is
`F_(2)=F-Ma=F=(MF)/(m+M)=(mF)/(m+F)` towards right
The situation from this frame is shown in figure. As the centre of mass is at rest in this frame, the block move in opposite direction and come to instantaneous rest at some instant. The extension of the spring will be maximum of this instant. Suppose the left block through a distance `x_(2)` from the initial position. 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)=(mF)/(m+F)(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,
`(mF)/(m+F)(x_(1)+x_(2))=(1)/(2)k(x_(1)+X_(2))^(2)`
or `x_(1)+x_(2)=(2mF)/(k(m+M))`
This the maximum extension of the spring.