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Q. A lift of mass $M =500\, kg$ is descending with speed of $2 \,ms ^{-1}$. Its supporting cable begins to slip thus allowing it to fall with a constant acceleration of $2\, ms ^{-2}$. The kinetic energy of the lift at the end of fall through to a distance of $6\, m$ will be ___ $kJ$.

JEE MainJEE Main 2023Work, Energy and Power

Solution:

$ v ^2 = u ^2+2 as $
$ =2^2+2(2)(6) $
$ =4+24=28$
$KE =\frac{1}{2} mv ^2 $
$ =\frac{1}{2}(500) 28 $
$ =7000 \,J $
$ =7\, kJ $