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Q.

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

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Detailed Solution

$u = 2 m / s; a = 2 m / s^{2}; s = 6 m$

$v^{2} = {(2)}^{2} + 2 \times 2 \times 6 = 28$

$K = \frac{1}{2} \times m V^{2} = 7 K J$

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A lift of mass M = 500 kg is descending with speed of 2 ms−1. Its supporting cable begins to slip thus allowing it of fall with a constant acceleration of 2 ms−2 . The kinetic energy of the lift at the end fall through to a distance of 6 m will be ________KJ.