A chain of mass 4 kg and length 2 m is hang | vedclass

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(D) The mass of the hanging part is $m = \frac{M}{n}$,where $M = 4 \ kg$ is the total mass and $n = 4$ is the fraction of the length hanging.So,$m = \

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$2.5$

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(D) The mass of the hanging part is $m = \frac{M}{n}$,where $M = 4 \ kg$ is the total mass and $n = 4$ is the fraction of the length hanging.So,$m = \frac{4}{4} = 1 \ kg$.The length of the hanging part is $l = \frac{L}{n} = \frac{2}{4} = 0.5 \ m$.The center of mass of the hanging part is at a distance $h = \frac{l}{2} = \frac{0.5}{2} = 0.25 \ m$ below the table edge.The work done to pull the hanging part onto the table is equal to the change in potential energy of the hanging part: $W = mgh$.Substituting the values: $W = 1 \ kg 	imes 10 \ m/s^2 	imes 0.25 \ m = 2.5 \ J$.

$A$ chain of mass $4 \ kg$ and length $2 \ m$ is hanging such that one-fourth of its length is hanging over the edge of a table. The work done to pull the hanging part onto the table is . . . . . . $J$. (Take $g = 10 \ m/s^2$)

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