In the figure,blocks A (mass 2m) and B (mass | vedclass

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(C) Before the string is cut,the system is in equilibrium. The spring force $F_s$ balances the total weight of the two blocks.$F_s = (m + 2m)g = 3mg$.

Vedclass · Accepted Answer

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

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(C) Before the string is cut,the system is in equilibrium. The spring force $F_s$ balances the total weight of the two blocks.$F_s = (m + 2m)g = 3mg$.Immediately after the string is cut,the spring force does not change instantaneously.For block $A$ (mass $2m$): The upward force is the spring force $F_s = 3mg$ and the downward force is its weight $2mg$. The net force $F_{net} = 3mg - 2mg = mg$ (upward).Acceleration $a_A = \frac{F_{net}}{2m} = \frac{mg}{2m} = \frac{g}{2}$ (upward).For block $B$ (mass $m$): The string is cut,so the tension becomes zero. The only force acting on block $B$ is its weight $mg$ (downward).Acceleration $a_B = \frac{mg}{m} = g$ (downward).Thus,the accelerations are $\frac{g}{2}$ and $g$ respectively.

In the figure,blocks $A$ (mass $2m$) and $B$ (mass $m$) are connected with a string and the system is suspended vertically with the help of a spring. The spring has negligible mass. Find the magnitude of the acceleration of masses $2m$ and $m$ just after the instant when the string is cut.

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