Thank you so much, you really did grasp my idea. Maybe I didn't ask the right questions, as you know sometimes it's about the quality of the question. Though I'm still trying to see how mechanical advantage does not alter the amount of power. Before we answer that, let me see if this rephrasing of my question will lead to a helpful answer and guide. Let's say I have a motor that has the below specification
MMA size
(shaft height, mm) | Rated Speed, rpm | Max Speed, rpm | Rated Power, kW | Rated Torque, N•m | Peak Torque, N•m |
|---|
| SH80 | 580…2600 | 1400…6000 | 0.6…8.6 | 10.5…31.5 | 21…88 |
| SH100 | 580…2600 | 1400…6000 | 2.3…21.1 | 35.6…94.6 | 80…200 |
| SH132 | 580…2600 | 1400…6000 | 6.4…48.4 | 90.7…217 | 200…400 |
If we attach 500kg of weight to the pulley weight to each of the motor and we attach a pulley with multiplying gears such that each rotation of the pulley in 2 seconds results to 100 rotation of the motor, which would translate to 30 rotation = 3,000 rpm. Now knowing the rpm, power output of the motor, peak torque and the height is 5 meters. Wouldn't it produce it produce the rated power as it's coming down?