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 .

Step-by-step solution

Given, u = 2   m   s - 1 ,   a = 2   m   s - 2 ,   s = 6   m Using kinematics third equation of motion, final speed of lift is v 2 = u 2 + 2 a s ⇒ v = 2 2 + 2 2 6 = 4 + 24 = 28   m   s - 1 Now, kinetic energy of the lift is K E = 1 2 m v 2 = 1 2 ( 500 ) 28 = 7000   J = 7   kJ