Question

) 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) The average useful power output of a person is 208.3 W

b) It take 141.14 s for this person to lift 2000 kg of bricks 1.50 m to a platform

Explanation:

a) We know that,

               $P=\frac{W}{t}$

Where ,

P is the power consumed  

W is the work done

t is the time elapsed

Given: t = 8 hr and   $W=6.00 \times 10^{6} J$

Substituting known information in the power equation, we get

             $P=\frac{6.00 \times 10^{6}}{8 h r}$

Convert hr into sec, $8 h r \times\left(\frac{3600 \text { sec }}{1 h r}\right)=28800 \mathrm{sec}$

So, $P=\frac{6.00 \times 10^{6}}{28800}=208.33 W$

b) Given:

m = 2000 kg

h = 1.50 m

we know, $g=9.8 \mathrm{m} / \mathrm{s}^{2}$

The formula for potential energy depends on the force acting on the two objects. For the gravitational force the formula is $W = m \times g \times h$, where m is the mass in kilograms, g is the acceleration due to gravity ($9.8 \mathrm{m} / \mathrm{s}^{2}$at the surface of the earth) and h is the height in meters.

Put this $W=m \times g \times h$in power equation, we get

        $P=\frac{m \times g \times h}{t}$

From above, find time by substituting the known values, we get

        $t=\frac{2000 \times 9.8 \times 1.50}{208.3}=\frac{29400}{208.3}=141.14 s$