A uniform chain of mass M and length L is pla | vedclass

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(C) The mass per unit length of the chain is $\lambda = \frac{M}{L}$.Initially,the length of the hanging part is $l_0 = \frac{L}{3}$. The mass of this

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$\frac{MgL}{18}$

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(C) The mass per unit length of the chain is $\lambda = \frac{M}{L}$.Initially,the length of the hanging part is $l_0 = \frac{L}{3}$. The mass of this hanging part is $m = \lambda l_0 = \frac{M}{L} \cdot \frac{L}{3} = \frac{M}{3}$.When a length $x$ of the chain is still hanging,the length of the hanging part is $(\frac{L}{3} - x)$.The weight of this hanging part is $F(x) = \lambda (\frac{L}{3} - x)g = \frac{M}{L}(\frac{L}{3} - x)g$.To pull the chain back onto the table,we must apply a force equal to this weight.The work done $W$ is the integral of the force over the displacement from $x = 0$ to $x = L/3$:$W = \int_{0}^{L/3} \frac{M}{L}(\frac{L}{3} - x)g \, dx$$W = \frac{Mg}{L} [\frac{L}{3}x - \frac{x^2}{2}]_{0}^{L/3}$$W = \frac{Mg}{L} [\frac{L^2}{9} - \frac{L^2}{18}] = \frac{Mg}{L} [\frac{L^2}{18}] = \frac{MgL}{18}$.

$A$ uniform chain of mass $M$ and length $L$ is placed on a frictionless table such that $1/3$ of its length is hanging over the edge. Find the work done to pull the hanging part back onto the table.

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