What is the power output of an electric motor that lifts a 2.0-kilogram block 15 meters vertically in 6.0 seconds?
1. 5.0 J
2. 5.0 W
3. 49 J
4. 49 W
1. 5.0 J
2. 5.0 W
3. 49 J
4. 49 W
-
power output
= work done / time
= 2 * 9.81 * 15 / 6
= 49.05W
answer
4. 49 W
= work done / time
= 2 * 9.81 * 15 / 6
= 49.05W
answer
4. 49 W
-
Power can be equal to either
= energy / time
= work / time
In this case, power = work / time
work = force * distance
In this case, the force = ma = mg, since the 2.0 kg block is being lifted up.
Ergo:
Force = 2.0 kg * 9.8 m/s^2 = 19.6 N
Work = 19.6 N * 15 m = 294 J
Power = 294 J / 6 s = 49 W
Answer: 4
= energy / time
= work / time
In this case, power = work / time
work = force * distance
In this case, the force = ma = mg, since the 2.0 kg block is being lifted up.
Ergo:
Force = 2.0 kg * 9.8 m/s^2 = 19.6 N
Work = 19.6 N * 15 m = 294 J
Power = 294 J / 6 s = 49 W
Answer: 4