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(B) Let the displacement of the plank relative to the ground be $x$ in the left direction. Since the system is on a smooth surface and there are no ex

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$0.6$

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(B) Let the displacement of the plank relative to the ground be $x$ in the left direction. Since the system is on a smooth surface and there are no external horizontal forces,the position of the center of mass of the system remains unchanged.Let the mass of the block be $m_1 = M$,the mass of the man be $m_2 = 2M$,and the mass of the plank be $m_3 = 2M$. The total mass of the system is $M_{total} = M + 2M + 2M = 5M$.The displacement of the center of mass is zero: $\Delta X_{cm} = \frac{m_1 \Delta x_1 + m_2 \Delta x_2 + m_3 \Delta x_3}{M_{total}} = 0$.Let the displacement of the plank be $x$ to the left $(-x)$.The man is on the plank,so his displacement is also $-x$.The block is initially $3 \ m$ away from the pulley. When it reaches the pulley,it has moved $3 \ m$ relative to the plank to the right. Its displacement relative to the ground is $\Delta x_1 = (3 - x)$ to the right.Substituting these values: $M(3 - x) + 2M(-x) + 2M(-x) = 0$.$3M - Mx - 2Mx - 2Mx = 0$.$3M = 5Mx$.$x = \frac{3}{5} = 0.6 \ m$.

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