Q 11-05-094JEE MainJEE Main 2023 (31 Jan, Shift 1)Easy
A lift of mass $M=500\ \text{kg}$ is descending with speed of $2\ \text{m s}^{-1}$. Its supporting cable begins to slip thus allowing it to fall with a constant acceleration of $2\ \text{m s}^{-2}$. The kinetic energy of the lift at the end of fall through to a distance of $6\ \text{m}$ will be ______ kJ.
Numerical value type. Enter your answer.
Answer: 7
$v^2=u^2+2as=4+2\times2\times6=28\ \text{m}^2\text{s}^{-2}$.
$K=\tfrac12\times500\times28=7000\ \text{J}=7\ \text{kJ}$.
Solution by Sreeraj P, M.Sc Physics