A rope ladder with a length l carrying a man | vedclass

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(A) Let the initial position of the balloon be $y_B = 0$ and the man be at $y_m = -l$. The center of mass of the system remains at the same vertical p

Vedclass · Accepted Answer

Increases by $mg(l - h)$

Vedclass · Answer

(A) Let the initial position of the balloon be $y_B = 0$ and the man be at $y_m = -l$. The center of mass of the system remains at the same vertical position because there are no external forces acting on the system.Initial center of mass $Y_{cm} = \frac{M(0) + m(-l)}{M + m} = \frac{-ml}{M + m}$.After the man climbs up,the balloon descends by $h$. The new position of the balloon is $y'_B = -h$ and the man is at $y'_m = -h$.New center of mass $Y'_{cm} = \frac{M(-h) + m(-h)}{M + m} = -h$.Equating $Y_{cm} = Y'_{cm}$,we get $\frac{-ml}{M + m} = -h$,which implies $ml = h(M + m) = Mh + mh$,so $Mh = m(l - h)$.The change in potential energy of the man is $\Delta PE_m = mg(y'_m - y_m) = mg(-h - (-l)) = mg(l - h)$.

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