Question

A uniform chain is held on a frictionless table with one-fifth of its length hanging
over the edge. If the chain has a length l and mass m, how much work is required to
pull the hanging part back on the table ?

A) mgl B) mgl/5 C) mgl/10 D) mgl/50

Answer 1

You can solve it by centre of mass. As hanging part is l/5.      Mass of l is =M                mass of l/5 is =m/5.           Work =P.E          h=l/10 .          Because Centre of mass of length l/5 /2=l/10.                                     W=mgh =mgl/50 Ans. I hope you like

Answer 2

Answer: D) mgl/50

Explanation:

Given:

$Mass of chain=M, Length of chain=L$

$Mass density per unit length =M/L$

As we know by the work-energy theorem, the work done is equal to the change in potential energy.

$W=P.E final-P.E initial$

The Centre of gravity for the hanging part is at:

$l/5×1/2=l/10$

l/5×1/2=l/10

The increase in potential energy will be :

$(m/5)×g×(l/10)=mgl/50$

(m/5)×g×(l/10)= mgl/50

Therefore, the Work done to move the hanging part of the chain back on the table will be mgl/50.