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

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

Vedclass · Accepted Answer

$3.6$

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(B) The total length of the chain is $L = 2\, m$ and its total mass is $M = 4\, kg$.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 edge of the table.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? ................ $J$

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