A uniform chain of length 2,m is kept on a ta | vedclass

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(B) Let the total length of the chain be $L = 2\,m$ and the total mass be $M = 4\,kg$.The length of the hanging part is $l = 60\,cm = 0.6\,m$.The mass

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$3.6$

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(B) Let the total length of the chain be $L = 2\,m$ and the total mass be $M = 4\,kg$.The length of the hanging part is $l = 60\,cm = 0.6\,m$.The mass per unit length of the chain is $\lambda = \frac{M}{L} = \frac{4}{2} = 2\,kg/m$.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 $h = \frac{l}{2} = \frac{0.6}{2} = 0.3\,m$ below the table edge.The work done to pull the chain onto the table is equal to the change in potential energy of the hanging part,which is $W = mgh$.Substituting the values: $W = 1.2 	imes 10 	imes 0.3 = 3.6\,J$.

$A$ uniform chain of length $2\,m$ is kept on a table such that a length of $60\,cm$ 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 entire chain on the table? (Take $g = 10\,m/s^2$)

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