Work-Energy Theorem question from JEE Main 2014 (19 Apr, Shift 1)

Answer: (a)

1. Each plank removes the same energy $Fd$. After one plank $v' = v\left(1 - \frac{1}{n}\right)$.
2. KE lost per plank $= \frac{1}{2}mv^2\left[1 - \left(1 - \frac{1}{n}\right)^2\right] = \frac{1}{2}mv^2 \cdot \frac{2n - 1}{n^2}$
3. Number of planks $= \frac{\frac{1}{2}mv^2}{\text{KE lost per plank}} = \frac{n^2}{2n - 1}$
4. So the answer is (a) $\frac{n^2}{2n-1}$.