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(B) Let the maximum extension in the spring be $x_{max}$. At maximum extension,the velocities of both blocks $M$ and $2M$ are zero in the center of ma

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$8F / 3K$

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(B) Let the maximum extension in the spring be $x_{max}$. At maximum extension,the velocities of both blocks $M$ and $2M$ are zero in the center of mass frame.Let $x_1$ and $x_2$ be the displacements of blocks $2M$ and $M$ respectively from their initial positions.From the work-energy theorem in the center of mass frame,the work done by the external forces equals the change in potential energy of the spring:$W_{ext} = \Delta U_{spring}$$F(x_1 + x_2) = \frac{1}{2} K x_{max}^2$,where $x_{max} = x_1 + x_2$.From the property of the center of mass,for the system to be at rest in the $CM$ frame,the displacements are inversely proportional to the masses:$2M x_1 = M x_2 \implies x_2 = 2x_1$.Since $x_{max} = x_1 + x_2 = x_1 + 2x_1 = 3x_1$,we have $x_1 = \frac{x_{max}}{3}$ and $x_2 = \frac{2x_{max}}{3}$.Substituting these into the work equation:$F(x_{max}) = \frac{1}{2} K x_{max}^2$However,considering the forces acting on the blocks: block $2M$ is pulled by a force $2F$ and block $M$ is pulled by a force $F$.The work done is $W = (2F)x_1 + (F)x_2 = 2F(\frac{x_{max}}{3}) + F(\frac{2x_{max}}{3}) = \frac{4F x_{max}}{3}$.Equating work done to potential energy: $\frac{4F x_{max}}{3} = \frac{1}{2} K x_{max}^2$.Solving for $x_{max}$: $x_{max} = \frac{8F}{3K}$.

In the diagram shown,there is no friction at any contact surface. Initially,the spring has no deformation. What will be the maximum deformation in the spring? Consider all the strings to be sufficiently large. Consider the spring constant to be $K$.

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