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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 to fall with a constant acceleration of $2 m s^{- 2}$ . The kinetic energy of the lift at the end of fall through to a distance of 6 m will be _______kJ.

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answer is 7.

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Detailed Solution

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

From equation of motion

$v^{2} = u^{2} + 2 as$

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

Kinetic energy K at the end of the fall is

$K = \frac{1}{2} \times {mv}^{2} = 7 KJ$

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