JEE Main 2024 — Work Power Energy Question with Solution

Given the length of the pendulum is $\ell =10 \mathrm{m}$.

The initial potential energy of the pendulum at the horizontal position is given by

$U=mg\ell ...\left(1\right)$

The final kinetic energy of the pendulum at the lowermost position is given by

$K=\frac{1}{2}m{v}^2 ...\left(2\right)$

So, from equations (1) and (2), it follows that

$$\begin{gathered} \frac{9}{10}mg\ell =\frac{1}{2}m{v}^2 \\ \Rightarrow \frac{9}{10}\times 10\times 10=\frac{1}{2}{v}^2 \\ \Rightarrow v^2=180 \\ \Rightarrow v=\sqrt{180} \\ =6\sqrt{5} \mathrm{m} {\mathrm{s}}^{-1} \end{gathered}$$