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(C) According to the Work-Energy Theorem,the work done by all forces (internal and external) on a system is equal to the change in its kinetic energy.

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(C) According to the Work-Energy Theorem,the work done by all forces (internal and external) on a system is equal to the change in its kinetic energy.$W_{total} = \Delta K = K_f - K_i$Here,the initial kinetic energy of the system is $K_i = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2 = 5 	imes (2)^2 = 20\,J$.The final kinetic energy of the system is $K_f = 0$ (since the blocks come to rest).Thus,$W_{total} = 0 - 20 = -20\,J$.However,the question asks for the work done by external forces only.On a frictionless horizontal surface,the external forces acting on the system are gravity and the normal force from the surface.Since these forces act perpendicular to the displacement of the center of mass (which is zero in this case as the system comes to rest),the work done by these external forces is $0\,J$.

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