Question

You are visiting your friend Fabio's house. You find that, as a joke, he filled his swimming pool with Kool-Aid, which dissolved perfectly into the water. However, now that you want to swim, you must remove all of the Kool-Aid contaminated water. The swimming pool is round, with a 9 foot radius. It is 8 feet tall and has 7.5 feet of water in it. How much work is required to remove all of the water by pumping it over the side? Use the physical definition of work, and the fact that the density of the Kool-Aid contaminated water is σ=63.5lbs/ft3

Answer 1

Answer:

Explanation:

The volume of contaminated water

= cross sectional area x height of water level

3.14 x 9 x 9 x 7.5 ft³

= 1907.55 ft³

mass = density x volume

= 1907.55 x 63.5 lbs

m = 121129.425 lbs

This mass has to be raised to the height of 8 ft before evacuation .

There is a rise of centre of mass of

8 - 7.5/2 ft

h = 4.25 ft

Energy required

= mgh

= 121129.425 x 32 x 4.25

= 16473601.8 unit.

Answer 2

Answer:

Workdone= 8.08×10^3Joules

Explanation:

dV= pir^2 dx

dV = pi×9^2 = 81pi

dF= density× pi× r^2 dx

dF= 63.5 × 81pi dx= 5143.5pi dx

W= Integral from7.5ft of water in the pool(8 -×)dx

W= 5143.5pi(8-7.5)

W= 5143.5 × 3.142 × 0.5

W= 8080.44Joules = 8.08×10^3Joules