An elevator can carry a maximum load of `1800 kg` (elevator + passengers) is moving up with a constant speed of `2 ms^(-1)`. The friction force opposite the motion is `4000 N`.What is minimum power delivered by the motor to the elevator?

Explanation

To solve the problem of finding the minimum power delivered by the motor to the elevator, we can follow these steps:

### Step 1: Identify the given values
- Maximum load of the elevator (mass, m) = 1800 kg
- Constant speed of the elevator (v) = 2 m/s
- Friction force (f_friction) = 4000 N
- Acceleration due to gravity (g) = 10 m/s² (approximately)

...