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 $kJ$.
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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
