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

The work done is equal to the increase in potential energy of the hanging part.Hanging length $L_h=60\text{cm}=0.6\text{m}$.Hanging mass $m_h=\frac{\text{Total mass}}{\text{Total length}}\times L_h=\frac{4}{2}\times 0.6=1.2\text{kg}$.The center of mass of the hanging part is at a distance $h=L_h/2=0.3\text{m}$ below the table.Work done $W=m_{h}gh$:
$$W=1.2\times 10\times 0.3=3.6\text{J}$$
(Taking $g=10{\text{m/s}}^2$ as per typical exam rounding or $g=9.8$ for $3.528\text{J}\approx 3.6\text{J}$)