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.
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CHECK THE SOLUTION.
(7)
v 2 = u 2 + 2as
= 2 2 + 2 (6) = 4 + 24 = 28
= 
(500)28
= 7 kJ
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