Question

A small electric motor is used to lift a 0.50-kilogram mass at constant speed. If the mass is lifted a vertical distance of 1.5 meters in 5.0 seconds, the average power developed by the motor is

Answer 1

Power developed by the motor is 1.47 W

Step-by-step explanation:

As we know that mechanical power is defined as the ratio of work done per unit time

here we know that work done to lift the mass upward is given as

`W = mgh`

here we have

m = 0.50 kg

h = 1.5 m

so we have

`W = 0.50(9.81)(1.5)`

`W = 7.36 J`

now it took 5 s to lift the mass so we have

`P = (W)/(t)`

`P = (7.36)/(5)`

`P = 1.47 W`