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

please briefly explain
( irrelevants will be reported)
​

Answer 1

Given :-

Initially, the mass of the chain = m

Length of the chain = l

The man drops the chain when half of the chain is in vertical position.

Solution :-

As the half of the chain is dropped, then, Centre of mass of chain = The part dropped down = $\dfrac{l}{2}$. Now, potential energy at the point is, P.E = mg$\dfrac{l}{2}$ - (a)

Centre of mass of chain = The part dropped down = $\dfrac{l}{4}$. Now, potential energy at the point is,

P.E' = $\dfrac{m}{2}$g$\dfrac{l}{4}$ - (b)

Now, Work done, W = Change in potential energy.

• W = P.E' - P.E

• 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 = $\dfrac{-3mgl}{8}$

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