Question

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

Answer 1

Power = 49 W

Step-by-step explanation:

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

Joule or J = kg*
`m^(2)`/
`s^(2)`

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*
`m^(2)`/
`s^(2)` = 294.3 J

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