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

Answer:

(a) 208.33 J/s

(b) 141.12 s

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 = $\frac{6 ×10^{6} }{28800}$

                                     = 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

                  = $\frac{29400}{208.33}$

                  = 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.