Two blocks A and B of masses 3,m and m respec | vedclass

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(C) Before the string is cut,the system is in equilibrium. For block $B$: The tension $T$ in the string is equal to the weight of block $B$,so $T = mg

Vedclass · Accepted Answer

$\frac{g}{3}, g$

Vedclass · Answer

(C) Before the string is cut,the system is in equilibrium. For block $B$: The tension $T$ in the string is equal to the weight of block $B$,so $T = mg$. For block $A$: The spring force $kx$ balances the weight of both blocks $A$ and $B$ plus the tension $T$. However,looking at the free body diagram of $A$,$kx = T + 3mg = mg + 3mg = 4mg$. Immediately after the string is cut,the tension $T$ becomes zero,but the spring force $kx$ remains $4mg$ because the spring does not change its length instantaneously. For block $B$: The only force acting on it is gravity,so its acceleration is $a_B = g$ (downwards). For block $A$: The net force is $F_{net} = kx - 3mg = 4mg - 3mg = mg$ (upwards). Therefore,the acceleration of $A$ is $a_A = \frac{F_{net}}{3m} = \frac{mg}{3m} = \frac{g}{3}$ (upwards). Thus,the accelerations of $A$ and $B$ are $\frac{g}{3}$ and $g$ respectively.

Two blocks $A$ and $B$ of masses $3\,m$ and $m$ respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in the figure. The magnitudes of acceleration of $A$ and $B$ immediately after the string is cut are respectively:

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