Question

(a) What is the average useful power output of a person who does 6.00×106 J of 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

(a) 208.33 J/s

(b) 141.12 s

Step-by-step explanation:

(a) Energy is the rate of doing work.

Work is the product of power and time

Work = Power × Time

Power = Work/time

where work is in joule and time is in seconds

Given that work = 6 ×
`10^(6)`

Time = 8 h = 8 × 60 × 60

= 28800 seconds

Average useful power =
`(6 ×10^(6) )/(28800)`

= 208.33 J/s

(b) Given

mass = 2000kg , using g = 9.8 m/s2

Force = 2000 × 9.8

= 19600N

Work = Force × Distance ( considering that Work done to lift his body can be omitted because it is not considered useful output here)

Work = 19600 × 1.5

= 29,400J

Time taken = Work/Power

=
`(29400)/(208.33) \\`

= 141.12s

≈ 2 Minutes 21 Seconds

At the rate of 208.33J/s, it will take the person 141.12 seconds to lift 2000 kg of bricks 1.50 m to a platform.