What is the power output of an electric motor that
lifts a 2.0-kilogram block 15 meters vertically in 6.0
seconds? PLEASE SHOW WORK
1. 5.0 J
2. 5.0 W
3. 49 J
4. 49 W

Respuesta :

Answer:

Power = 49 W

Explanation:

power = work / time (J/t or watt)

Joule or J = kg*[tex]m^{2}[/tex]/[tex]s^{2}[/tex]

watt or W = J/s

given:

mass = 2.0 kg

height = 15 m

g = 9.81 m/s2  (assumed)

t = 6 seconds

work = mass*g*h

work = 2.0*9.81*15 = 294.3 kg*[tex]m^{2}[/tex]/[tex]s^{2}[/tex] = 294.3 J

Power = 294.3 J / 6 seconds = 49.1 J/s = 49 W

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