A uniform chain of length 3 m and mass 3 kg overhangs a smooth table with 2 m laying on the table. If k is the kinetic energy of the chain in joule as it completely slips off the table, then the value of k is ……… .

(Take, g = 10 m/s²)

The given situation is shown below as

Applying law of conservation of energy, we get

$\begin{array}{l}{\mathrm{K}}_{\text{initial }}+{\mathrm{U}}_{\text{initial }}={\mathrm{K}}_{\text{final }}+{\mathrm{U}}_{\text{final }}\\ {\mathrm{K}}_{\text{initial }}=0\\ \Rightarrow {\mathrm{U}}_{\text{initial }}=-1\times 10\times \frac{1}{2}\\ {\mathrm{U}}_{\text{final }}=-3\times 10\times \frac{3}{2}(3\mathrm{m}\text{ lie below DATUM level, so negative)}\\ 0-1\times 10\times \frac{1}{2}={\mathrm{K}}_{\text{final }}-3\times 10\times \frac{3}{2}\\ \Rightarrow -5={\mathrm{K}}_{\text{final }}-45\\ \Rightarrow {\mathrm{K}}_{\text{final }}=45-5=40 \mathrm{J}\end{array}$