Question

A uniform chain of mass m and length l overhangs a table with its two third parts on the table. Find the work to be done by a person to put the hanging part back on the table.

Answer 1

The mass is uniformly distributed. Two third parts are on the table, so one third part is hanging. Since one third of the hanging mass of the hanging part is $\dfrac{m}{3}$ and this weight will be at the centre of mass of the hanging part.

Complete step by step answer:
Now the mass of the hanging part is $\dfrac{m}{3}$, we can use the work energy theorem here to solve this problem.
Total length is l and the length of the hanging part will be $\dfrac{l}{3}$. Since, the mass distribution is uniform, the centre of mass of this hanging part will be at the middle point that is at $\dfrac{l}{6}$.
Potential energy of this hanging part $=mgh$
$\Rightarrow \dfrac{m}{3}\times g\times \dfrac{l}{6}$
$\Rightarrow \dfrac{mgl}{18}$
This is the potential energy. Work energy theorem states that work done is equal to the change in the potential energy of the object. So, when this hanging part goes up, its potential energy becomes zero.
Thus, the work to be done by a person to put the hanging part back on the table is $\dfrac{mgl}{18}$ Joules.

Note: Work energy theorem states that work done is equal to the change in the potential energy of the object. The rope has a uniform mass m. So, the centre of mass for the rope is at half of the length of the rope.