Question:
A uniform chain of length 3 meter and mass $3 \mathrm{~kg}$ overhangs a smooth table with 2 meter laying on the table. If $\mathrm{k}$ is the kinetic energy of the chain in joule as it completely slips off the table, then the value of $k$ is
Solution:
From energy conservation
$\mathrm{K}_{\mathrm{i}}+\mathrm{U}_{\mathrm{i}}=\mathrm{k}_{\mathrm{f}}+\mathrm{U}_{\mathrm{f}}$
$0+\left(-1 \times 10 \times \frac{1}{2}\right)=\mathrm{k}_{\mathrm{f}}+\left(-3 \times 10 \times \frac{3}{2}\right)$
$-5=k_{f}-45$
$k_{f}=40 \mathrm{~J}$
Ans. $40.00$
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