What is the average useful power output of a person who does 5.50×106 J of useful work in 6.50 h?
Working at this rate, how long will it take this person to lift 1850 kg of bricks 1.20 m to a platform? (Work done to lift his body can be omitted because it is not considered useful output here.)
A. t = 6.50h * 3600s/h = 23,400 s.
Po = W/t = 5.50*10^6J/2.34*10^4s = 235 J/s = 235 Watts.
B. Work = F*d = Mg*d = 1850*9.8 * 1.2 =
21,756 J.
t = 21,756J./(235J/s) = 92.6 s.
To find the average useful power output, we need to divide the useful work done by the time taken.
1) Calculate the average useful power output:
Average power = Useful work / Time
Given that the person does 5.50×10^6 J of useful work in 6.50 h, we can calculate the average power by converting the time to seconds:
Time in seconds = 6.50 h * 60 min/h * 60 s/min = 23,400 s
Average power = 5.50×10^6 J / 23,400 s
2) Calculate the time it takes to lift 1850 kg of bricks 1.20 m:
To solve this, we need to use the equation for work:
Work = Force * Distance
Given that we want to lift 1850 kg of bricks 1.20 m, we know the work done is equal to:
Work = (mass of bricks) * (acceleration due to gravity) * (height)
Work = 1850 kg * 9.8 m/s^2 * 1.20 m
Work = 21,672 J
Now we can use the formula for power to determine the time it takes:
Power = Work / Time
Plug in the values we know:
Power = 21,672 J / (average power output in watts from step 1)
By rearranging the formula, we can solve for time:
Time = Work / Power
Substituting the values:
Time = 21,672 J / (5.50×10^6 J / 23,400 s)
Now we can calculate the time it takes to lift the bricks.