JEE MAIN - Physics (2023 - 31st January Morning Shift - No. 18)
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
7
Explanation
Given, $u=2 \mathrm{~m} / \mathrm{s}$
$$ \begin{aligned} & a=2 \mathrm{~m} / \mathrm{s}^{2} \\\\ & s=6 \mathrm{~m} \\\\ & v=? \\\\ & v^{2}=u^{2}+2 a s \\\\ & v^{2}=4+2 \times 2 \times 6 \\\\ & =28 \end{aligned} $$
So, $\mathrm{KE}=\frac{1}{2} m v^{2}=\frac{1}{2} \times 500 \times 28 \mathrm{~J}$
$=7000 \mathrm{~J}$
$=7 \mathrm{~kJ}$
$$ \begin{aligned} & a=2 \mathrm{~m} / \mathrm{s}^{2} \\\\ & s=6 \mathrm{~m} \\\\ & v=? \\\\ & v^{2}=u^{2}+2 a s \\\\ & v^{2}=4+2 \times 2 \times 6 \\\\ & =28 \end{aligned} $$
So, $\mathrm{KE}=\frac{1}{2} m v^{2}=\frac{1}{2} \times 500 \times 28 \mathrm{~J}$
$=7000 \mathrm{~J}$
$=7 \mathrm{~kJ}$
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