A lift of mass M = 500 kg is descending with speed of 2 ms⁻¹. Its supporting cable begins to slip thus allowing it to fall with a constant acceleration of 2 ms⁻². The kinetic energy of the lift at the end of fall through to a distance of 6 m will be _____________ kJ. Numerical answer type. Work it out, then open the answer.

Answer: 7. v² = u² + 2as = 2² + 2(2)(6) = 28 m²/s² K = 1/2Mv² = 1/2(500)(28) = 7000 J So K = 7 kJ.

A lift of mass M = 500 kg is descending with speed of 2 ms⁻¹. Its… | JEE Main 2023 (31 Jan, Shift 1)

Q9JEE Main 2023 · 31 JAN · SHIFT 1Medium

Work-Energy Theorem

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

Numerical answer type. Work it out, then open the answer.

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Answer: 7

$v^2 = u^2 + 2as = 2^2 + 2(2)(6) = 28\text{ m}^2/\text{s}^2$
$K = \frac{1}{2}Mv^2 = \frac{1}{2}(500)(28) = 7000\text{ J}$
So $K = 7\text{ kJ}$ .

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Work-Energy Theorem question from JEE Main 2023 (31 Jan, Shift 1)

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