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(B) The length of the chain is $L = 2 \ m$ and the total mass is $M = 4 \ kg$. The mass per unit length is $\lambda = M/L = 4/2 = 2 \ kg/m$.The length

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$3.6$

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(B) The length of the chain is $L = 2 \ m$ and the total mass is $M = 4 \ kg$. The mass per unit length is $\lambda = M/L = 4/2 = 2 \ kg/m$.The length of the hanging part is $l = 60 \ cm = 0.6 \ m$.The mass of the hanging part is $m = \lambda 	imes l = 2 	imes 0.6 = 1.2 \ kg$.The center of mass of the hanging part is at a distance $h = l/2 = 0.6/2 = 0.3 \ m$ below the edge of the table.To pull the chain onto the table,we must lift the center of mass of the hanging part to the level of the table.The work done $W$ is equal to the change in potential energy of the hanging part: $W = mgh$.$W = 1.2 \ kg 	imes 10 \ m/s^2 	imes 0.3 \ m = 3.6 \ J$.

$A$ uniform chain of length $2 \ m$ is placed on a table such that $60 \ cm$ of it hangs freely from the edge of the table. The total mass of the chain is $4 \ kg$. What is the work done in pulling the chain completely onto the table (in $J$)? (Take $g = 10 \ m/s^2$)

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