A lift of mass M = 500   kg is descending with speed of 2   m   s - 1 . Its supporting cable…

Given, $u=2 \mathrm{m} {\mathrm{s}}^{-1}, a=2 \mathrm{m} {\mathrm{s}}^{-2}, s=6 \mathrm{m}$

Using kinematics third equation of motion, final speed of lift is $v^2=u^2+2as$

$\Rightarrow v=2^2+2\left(2\right)\left(6\right)$

$=4+24=28 \mathrm{m} {\mathrm{s}}^{-1}$

Now, kinetic energy of the lift is $KE=\frac{1}{2}m{v}^2$

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

$=7000 \mathrm{J}$

$=7 \mathrm{kJ}$