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