A lift of mass $\mathrm{M}=500 \mathrm{~kg}$ is descending with speed of $2 \mathrm{~ms}^{-1}$. Its supporting cable begins to slip thus allowing it to fall with a constant acceleration of $2 \mathrm{~ms}^{-2}$. The kinetic energy of the lift at the end of fall through to a distance of $6 \mathrm{~m}$ will be _____________ $\mathrm{kJ}$.
Answer (integer)
7
Solution
Given, $u=2 \mathrm{~m} / \mathrm{s}$
<br/><br/>$$
\begin{aligned}
& a=2 \mathrm{~m} / \mathrm{s}^{2} \\\\
& s=6 \mathrm{~m} \\\\
& v=? \\\\
& v^{2}=u^{2}+2 a s \\\\
& v^{2}=4+2 \times 2 \times 6 \\\\
& =28
\end{aligned}
$$
<br/><br/>So, $\mathrm{KE}=\frac{1}{2} m v^{2}=\frac{1}{2} \times 500 \times 28 \mathrm{~J}$
<br/><br/>$=7000 \mathrm{~J}$
<br/><br/>$=7 \mathrm{~kJ}$
About this question
Subject: Physics · Chapter: Work, Energy and Power · Topic: Work Done by a Force
This question is part of PrepWiser's free JEE Main question bank. 80 more solved questions on Work, Energy and Power are available — start with the harder ones if your accuracy is >70%.