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

We need to find the minimum power delivered by the motor of an elevator carrying a maximum load of 1400 kg, moving upward with a uniform speed of 3 m/s, against a frictional force of 2000 N.

We begin by determining the total weight of the system. The total mass is 1400 kg (600 kg passengers + 800 kg elevator). Taking $$g = 10$$ m/s$$^2$$ gives:
$$W = mg = 1400 \times 10 = 14000 \text{ N}$$

Next, we determine the total force the motor must exert. Since the elevator moves upward at uniform speed, the acceleration is zero ($$a = 0$$), so by Newton's second law the net force is zero and the motor must exert an upward force equal to the sum of the weight (downward) and friction (opposing motion):
$$F_{motor} = W + F_{friction} = 14000 + 2000 = 16000 \text{ N}$$

Then, using the power formula for a constant force applied in the direction of motion at constant velocity, namely $$P = F \times v$$, where $$F$$ is the force and $$v$$ is the velocity, and substituting in the values gives:
$$P = 16000 \times 3 = 48000 \text{ W} = 48 \text{ kW}$$

Therefore, the minimum power delivered by the motor is 48 kW.