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(B) Before the string breaks,the system is in equilibrium.For the lower block of mass $m$,the spring force $F_s = kx = mg$ (upward).For the upper bloc

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$3\,g / 2$

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(B) Before the string breaks,the system is in equilibrium.For the lower block of mass $m$,the spring force $F_s = kx = mg$ (upward).For the upper block of mass $2\,m$,the tension $T$ in the string balances the weight and the spring force: $T = 2mg + F_s = 2mg + mg = 3mg$.Immediately after the string breaks,the tension $T$ becomes $0$,but the spring force $F_s$ remains $mg$ (downward on the upper block) because the spring cannot change its length instantaneously.Applying Newton's second law to the upper block of mass $2\,m$:$F_{net} = (2m)g + F_s = 2mg + mg = 3mg$.$a = F_{net} / (2m) = 3mg / 2m = 3g / 2$.

Two blocks are connected by a spring. The combination is suspended,at rest,from a string attached to the ceiling,as shown in the figure. The string breaks suddenly. Immediately after the string breaks,what is the initial downward acceleration of the upper block of mass $2\,m$?

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