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.

Solution :

Given :

Mass of lift,

$$M = 500\text{ kg}$$

Initial speed,

$$u = 2\text{ m s}^{-1}$$

Acceleration downward,

$$a = 2\text{ m s}^{-2}$$

Distance fallen,

$$s = 6\text{ m}$$

Using equation of motion :

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

$$= (2)^2 + 2(2)(6)$$

$$= 4 + 24$$

$$= 28$$

Kinetic energy at the end :

$$K = \frac{1}{2}Mv^2$$

$$= \frac{1}{2}(500)(28)$$

$$= 250 \times 28$$

$$= 7000\text{ J}$$

$$= 7\text{ kJ}$$

Final Answer :

$$7\text{ kJ}$$