The system shown in the figure is in equilibr | vedclass

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(A) Initially,both the blocks are in equilibrium. For mass $m$:$T = mg$For mass $2m$:$F_s = T + 2mg = mg + 2mg = 3mg$Immediately after the string is c

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$g/2$ upwards,$g$ downwards

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(A) Initially,both the blocks are in equilibrium. For mass $m$:$T = mg$For mass $2m$:$F_s = T + 2mg = mg + 2mg = 3mg$Immediately after the string is cut,the tension $T$ becomes $0$,but the spring force $F_s$ remains $3mg$ because the spring does not change its length instantaneously.For mass $m$:The only force acting is gravity ($mg$ downwards).$mg = ma_m$$a_m = g$ (downwards)For mass $2m$:The forces are $F_s$ (upwards) and $2mg$ (downwards).$F_s - 2mg = (2m)a_{2m}$$3mg - 2mg = 2ma_{2m}$$mg = 2ma_{2m}$$a_{2m} = g/2$ (upwards)Thus,the acceleration of mass $2m$ is $g/2$ upwards and mass $m$ is $g$ downwards.

The system shown in the figure is in equilibrium and at rest. The spring and string are massless. Now,the string is cut. The acceleration of mass $2m$ and $m$ just after the string is cut will be:

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