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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

Respuesta :

Answer:

Power developed by the motor is 1.47 W

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

[tex]W = mgh[/tex]

here we have

m = 0.50 kg

h = 1.5 m

so we have

[tex]W = 0.50(9.81)(1.5)[/tex]

[tex]W = 7.36 J[/tex]

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

[tex]P = \frac{W}{t}[/tex]

[tex]P = \frac{7.36}{5}[/tex]

[tex]P = 1.47 W[/tex]

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