Work Energy Power

JEE Main level, numerical answer, medium

75 of 183

Question 75 of 183

A lift of mass M=500kgM=500kg is descending with speed of 2ms−12m{s}^{-1}. Its supporting cable begins to slip thus allowing it to fall with a constant acceleration of 2ms−22m{s}^{-2}. The kinetic energy of the lift at the end of fall through to a distance of 6m6m will be ______ kJkJ.

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Answer: 7

Given, u=2ms−1,a=2ms−2,s=6mu=2m{s}^{-1},a=2m{s}^{-2},s=6m
Using kinematics third equation of motion, final speed of lift is v2=u2+2as{v}^{2}={u}^{2}+2as
⇒v=22+2(2)(6)\Rightarrow v={2}^{2}+2\left(2\right)\left(6\right)
=4+24=28ms−1=4+24=28m{s}^{-1}
Now, kinetic energy of the lift is KE=12mv2KE=\dfrac{1}{2}m{v}^{2}
=12(500)28=\dfrac{1}{2}(500)28
=7000J=7000J
=7kJ=7kJ

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