A uniform chain of length L = 2,m is kept on | vedclass

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(C) The mass per unit length of the chain is $\lambda = \frac{M}{L} = \frac{4\,kg}{2\,m} = 2\,kg/m$.The mass of the hanging part is $m = \lambda 	ime

Vedclass · Accepted Answer

$3.6$

Vedclass · Answer

(C) The mass per unit length of the chain is $\lambda = \frac{M}{L} = \frac{4\,kg}{2\,m} = 2\,kg/m$.The mass of the hanging part is $m = \lambda 	imes \ell = 2\,kg/m 	imes 0.6\,m = 1.2\,kg$.The center of mass of the hanging part is at a distance of $h = \frac{\ell}{2} = \frac{0.6\,m}{2} = 0.3\,m$ below the edge of the table.The work done $W$ to pull the hanging part onto the table is equal to the change in potential energy of the hanging part,which is $W = mgh = (1.2\,kg) 	imes (10\,m/s^2) 	imes (0.3\,m) = 3.6\,J$.

$A$ uniform chain of length $L = 2\,m$ is kept on a table such that a length $\ell = 60\,cm$ hangs freely from the edge of the table. The total mass of the chain is $M = 4\,kg$. The work done in pulling the entire chain onto the table is ............. $J$. (Take $g = 10\,m/s^2$)

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