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…

Question

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.

Accepted Answer

Correct answer: 7. (7) v^2 = u^2 + 2as = 2^2 + 2 (b)(6) = 4 + 24 = 28 K E= 12m v^2 = 12 (500)28 = 7 kJ

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

Solution

More Newton's Laws of Motion PYQs