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(B) Let the mass of each block be $m$. The system moves with a common acceleration $a = \frac{F}{2m}$.In the frame of the centre of mass, the blocks a

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$10.4$

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(B) Let the mass of each block be $m$. The system moves with a common acceleration $a = \frac{F}{2m}$.In the frame of the centre of mass, the blocks are at rest at the moment of maximum extension. The pseudo force acting on each mass is $F_p = ma = \frac{F}{2}$.Let $x_1$ and $x_2$ be the displacements of the two blocks from their initial positions in the centre of mass frame. The total extension of the spring is $x = x_1 + x_2$.The work done by the external force and pseudo forces in the centre of mass frame is equal to the change in potential energy of the spring:$W = \frac{1}{2} k x^2$In the centre of mass frame, the work done by the effective forces is:$(F - ma)x_1 + (ma)x_2 = \frac{1}{2} k (x_1 + x_2)^2$Substituting $a = \frac{F}{2m}$:$(F - \frac{F}{2})x_1 + (\frac{F}{2})x_2 = \frac{1}{2} k (x_1 + x_2)^2$$\frac{F}{2}(x_1 + x_2) = \frac{1}{2} k (x_1 + x_2)^2$$x_1 + x_2 = \frac{F}{k}$Given $F = 10 \,N$ and $k = 2500 \,N/m$, the maximum extension is:$x_{max} = \frac{10}{2500} \,m = 0.004 \,m = 0.4 \,cm$The maximum distance between the blocks is the sum of the natural length and the maximum extension:$d_{max} = 10 \,cm + 0.4 \,cm = 10.4 \,cm$.

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