Given, u=2ms−1,a=2ms−2,s=6m
Using kinematics third equation of motion, final speed of lift is v2=u2+2as
⇒v=22+2(2)(6)
=4+24=28ms−1
Now, kinetic energy of the lift is KE=21mv2
=21(500)28
=7000J
=7kJ
A lift of mass M=500kg is descending with speed of 2ms−1. Its supporting cable begins to slip thus allowing it to fall with a constant acceleration of 2ms−2. The kinetic energy of the lift at the end of fall through to a distance of 6m will be ______ kJ.
Held on 31 Jan 2023 · Verified 6 Jul 2026.
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