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 $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and the frictional force acting on it is 2000 N , then the maximum power used by the motor is $\_\_\_\_$ $\mathrm { kW } . \quad g = 10 \mathrm {~m} \mathrm {~s} ^ { - 2 }$