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

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Solution

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 to fall with a constant acceleration of 2 m $s^{- 2}$ . The kinetic energy of the lift at the end of fall through a distance of 6 m will be ________ kJ. [2023]

Ans.
(7)

$v^{2} = u^{2} + 2 a s = 2^{2} + 2 (2) (6)$

$= 4 + 24 = 28$

$K E = \frac{1}{2} m v^{2} = \frac{1}{2} (500) (28) = 7000 J = 7 kJ$

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