Question

a) What is the average useful power output of a person who does6.00×10^6Jof useful work in 8.00 h? (b) Working at this rate, how long will it take this person to lift 2000 kg of bricks 1.50 m to a platform? (Work done to lift his body can be omitted because it is not considered useful output here

Answer 1

208.33 W

141.26626 seconds

Step-by-step explanation:

E = Energy =
`6* 10^6\ J`

t = Time taken = 8 h

m = Mass = 2000 kg

g = Acceleration due to gravity = 9.81 m/s²

h = Height of platform = 1.5 m

Power is obtained when we divide energy by time

`P=(E)/(t)\\\Rightarrow P=(6* 10^6)/(8* 60* 60)\\\Rightarrow P=208.33\ W`

The average useful power output of the person is 208.33 W

The energy in the next part would be the potential energy

The time taken would be

`t=(E)/(P)\\\Rightarrow t=(mgh)/(208.33)\\\Rightarrow t=(2000* 9.81* 1.5)/(208.33)\\\Rightarrow t=141.26626\ s`

The time taken to lift the load is 141.26626 seconds