For the system shown in the figure,m_1 m_2 | vedclass

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(B) Before cutting the string,the system is in equilibrium. The spring force does not change instantaneously when the string is cut. Therefore,the for

Vedclass · Accepted Answer

Only $I$,$II$ and $IV$ are correct

Vedclass · Answer

(B) Before cutting the string,the system is in equilibrium. The spring force does not change instantaneously when the string is cut. Therefore,the forces acting on $m_1$,$m_2$,and $m_3$ remain unchanged immediately after the string is cut. Thus,$m_1$,$m_2$,and $m_3$ remain stationary (Statement $I$ is correct).Since $m_1$,$m_2$,and $m_3$ are stationary,their acceleration is $0$. The block $m_4$ is now subject to the spring force (which was balancing its weight and the tension) and gravity. Thus,$m_4$ will accelerate (Statement $IV$ is correct).Since we can calculate the forces acting on all blocks,the acceleration of all $4$ blocks can be determined (Statement $II$ is correct).Therefore,statements $I$,$II$,and $IV$ are correct.

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