Verified Solution • 2025–26 Session moderate

QJEEM23Ph28ed: JEE Main 2023 Physics - Class 11 JEEM23Ph28ed Solution: A lift of mass is descending with speed of . Its supporting…

A lift of mass M=500 kg\mathrm{M}=500 \mathrm{~kg} is descending with speed of 2 ms−12 \mathrm{~ms}^{-1} . Its supporting cable begins to slip thus allowing it to fall with a constant acceleration of 2 ms−22 \mathrm{~ms}^{-2} . The kinetic energy of the lift at the end of fall through to a distance of 6 m6 \mathrm{~m} will be _____________ kJ\mathrm{kJ} .

Step-by-Step Verified Solution

Given, u=2 m/su=2 \mathrm{~m} / \mathrm{s}

amp;a=2 m/s2amp;s=6 mamp;v=?amp;v2=u2+2asamp;v2=4+2×2×6amp;=28\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, KE=12mv2=12×500×28 J\mathrm{KE}=\frac{1}{2} m v^{2}=\frac{1}{2} \times 500 \times 28 \mathrm{~J}

=7000 J=7000 \mathrm{~J}

=7 kJ=7 \mathrm{~kJ}

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