Question

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 ___ $kJ$

Answer 1

The correct answer is 7.
v2=u2+2as
=22+2(2)(6)
=4+24=28
KE=21​mv2
=21​(500)28
=7000J
=7kJ