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

<p>First, let&#39;s find the total weight of the elevator and passengers:</p> <p>Total weight = (mass of passengers + mass of elevator) × g<br/><br/> Total weight = (600 kg + 800 kg) × 10 m/s²<br/><br/> Total weight = 1400 kg × 10 m/s² = 14,000 N</p> <p>Now, we need to calculate the total force acting on the elevator as it moves upwards. Since the elevator is moving at a constant speed, the net force acting on it is zero. Therefore, the tension in the cable must balance the total weight and frictional force:</p> <p>Tension = Total weight + Frictional force Tension = 14,000 N + 2,000 N = 16,000 N</p> <p>The power used by the motor can be calculated using the formula:</p> <p>Power = Force × Velocity</p> <p>Here, the force is the tension in the cable, and the velocity is the speed of the elevator:</p> <p>Power = 16,000 N × 3 m/s = 48,000 W</p> <p>To convert the power to kilowatts, divide by 1,000:</p> <p>Power = 48,000 W / 1,000 = 48 kW</p> <p>So, the maximum power used by the motor is 48 kW.</p>