Question

A uniform chain of length L and mass M is lying on a smooth table and one-third of its length is
hanging vertically down over the edge of the table. Ifg is acceleration due to gravity, then work
required to pull the hanging part on the table is
a) Mgl
b) MgL/3
c) MgL/9
d) MgL/18
​

Answer 1

Answer:

d) MgL/18

Explanation:

Mass of chain on table= $\frac{2M}{3}$

Mass of chain hanging=$\frac{M}{3}$

Considering the table as the reference plane, CoM of hanging part is at $\frac{-l}{6}$

Center of mass of the chain initially is given by:

=$\frac{(2M/3)(0)+(M/3)(-l/6)}{M}$

= $\frac{-l}{18}$

∴ Initial gravitational potential energy =$-mg\frac{l}{18}$

And final G.P.E= $mg(0)=0$

∴ work done = $0-[-mg\frac{l}{18}]$ = $\boxed{\boxed{mg\frac{l}{18} }}$

 

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