A uniform chain of length 2 m is kept on a table such that a length of 60 cm hangs freely from the edge of the table. The total mass of the chain is 4 kg. What is the work done in pulling the entire chain on the table?

3.6 J

The total length of the chain = 2 m

The total mass of the chain = 4 kg

60 cm of the chain is hanging from the table

Fraction of length of the chain hanging from the table : \(\frac{1}{n}\)

\(\frac{1}{n} = \frac{60}{120} = \frac{3}{10}\)

\(\Rightarrow n = \frac{10}{3}\)

Now, Work done in pulling the chain on the table

As we know if we pull the chain, each part of the chain will not cover same height, therefore will not contribute the same potential energy, so let's take mass of the chain.

\(L' = \frac{L}{2n}\)

Thus, \(W = \frac{mgL}{2n^2}\)

\(W = \frac{4*10*2}{2*(10/3)^2}\)

\(W = 3.6 \text{ J}\)

Hence the work done in pulling the chain is 3.6 J.