The system shown in the figure is in equilibr | vedclass

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(A) Initially,the system is in equilibrium. The spring force $kx$ balances the total weight of the two masses.$kx = (2m + m)g = 3mg$Just after the str

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$\frac{g}{2}$ upwards,$g$ downwards

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(A) Initially,the system is in equilibrium. The spring force $kx$ balances the total weight of the two masses.$kx = (2m + m)g = 3mg$Just after the string is cut,the spring force remains unchanged because the spring cannot change its length instantaneously. The tension in the string becomes zero.For mass $m$ (lower block): The only force acting on it is gravity. Thus,its acceleration is $a_m = g$ downwards.For mass $2m$ (upper block): The forces acting on it are the spring force $kx$ upwards and gravity $2mg$ downwards.Using Newton's second law: $F_{	ext{net}} = ma$$kx - 2mg = (2m)a_{2m}$Substituting $kx = 3mg$:$3mg - 2mg = 2m \cdot a_{2m}$$mg = 2m \cdot a_{2m}$$a_{2m} = \frac{g}{2}$ upwards.Therefore,the acceleration of mass $2m$ is $\frac{g}{2}$ upwards and the acceleration of 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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