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(C) The system consists of three masses $m_1, m_2, m_3$ connected by a string.Mass $m_1$ is hanging vertically,so the driving force is its weight $m_1

Vedclass · Accepted Answer

$\frac{g(1 - 2\mu)}{3}$

Vedclass · Answer

(C) The system consists of three masses $m_1, m_2, m_3$ connected by a string.Mass $m_1$ is hanging vertically,so the driving force is its weight $m_1 g$.Masses $m_2$ and $m_3$ are on a rough horizontal surface,so the opposing frictional force is $f = \mu(m_2 + m_3)g$.The net force acting on the system is $F_{net} = m_1 g - \mu(m_2 + m_3)g$.The total mass of the system is $M_{total} = m_1 + m_2 + m_3$.Using Newton's second law,the acceleration $a$ is given by $a = \frac{F_{net}}{M_{total}} = \frac{m_1 g - \mu(m_2 + m_3)g}{m_1 + m_2 + m_3}$.Given $m_1 = m_2 = m_3 = m$,we substitute these values:$a = \frac{mg - \mu(m + m)g}{m + m + m} = \frac{mg - 2\mu mg}{3m} = \frac{mg(1 - 2\mu)}{3m} = \frac{g(1 - 2\mu)}{3}$.

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