### Laws Of Motion Question 11

#### Question 11 - 31 January - Shift 1

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 $kJ$.

## Show Answer

#### Answer: (7)

#### Solution:

$ \begin{aligned} v^{2} & =u^{2}+2 \text{ as } \\ & =2^{2}+2(2)(6) \\ & =4+24=28 \\ KE & =\frac{1}{2} mv^{2} \\ & =\frac{1}{2}(500) 28 \\ & =7000 J \\ & =7 kJ \end{aligned} $

Ans. 7