If the maximum load carried by an elevator is 1400 kg ( 600 kg - Passengers +800 kg - elevator), which is moving up with a uniform speed of 3 m s^-1…

If the maximum load carried by an elevator is $\displaystyle 1400$ kg ( $\displaystyle 600$ kg - Passengers +$\displaystyle 800$ kg - elevator), which is moving up with a uniform speed of $\displaystyle 3 \mathrm{~m} \mathrm{~s}^{-1}$ and the frictional force acting on it is $\displaystyle 2000$ N , then the maximum power used by the motor is $\displaystyle \_\_\_\_$ $\displaystyle \mathrm{kW}\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)$