Question 248: Work, Energy and Power & Circular Motion

Question

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

Accepted Answer

Given, $u=2 \mathrm{~m} / \mathrm{s}$
$$
\begin{aligned}
& a=2 \mathrm{~m} / \mathrm{s}^{2} \\
& s=6 \mathrm{~m} \\
& v=? \\
& v^{2}=u^{2}+2 a s \\
& v^{2}=4+2 	imes 2 	imes 6 \\
& =28
\end{aligned}
$$
So, $\mathrm{KE}=\frac{1}{2} m v^{2}=\frac{1}{2} 	imes 500 	imes 28 \mathrm{~J}$
$=7000 \mathrm{~J}$
$=7 \mathrm{~kJ}$

Work, Energy and Power & Circular Motion

Choose the correct answer: