Question

A man places a vertical chain (of mass 'm' and length 'C') on a table slowly. Initially the lower end of the
chain just touches the table. The man drops the chain when half of the chain is in vertical position. Then work done by the man in this process is:
(1)-mgl/2
(2)-mgl/4
(3)-3mgl/8
(4)-mgl/8

( irrelevants will be reported)
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Answer 1

Solution : Change in potential energy = Work done

› W = $\dfrac{m}{2}$g$\dfrac{l}{4}$ - mg$\dfrac{l}{2}$

› W = mg ($\dfrac{l}{8}$ - $\dfrac{l}{2}$)

› W = mg($\dfrac{-3l}{8}$)

› W = $\dfrac{-3mgl}{8}$

$\bold{Hope\;it \; helps\;!}$