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 18 foot radius. It is 11 feet tall and has 6 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=63.8lbs/ft^3. Don't forget to enter the correct units.

Answer 1

Answer:

The amount of work required to remove all of the water by pumping it over the side = 3.12 × 10⁶ lbs.ft

Step-by-step explanation:

Work done in moving anything from point A to point B = Fx

For this setup,

If we take an elemental vertical height, dx, The volume would be A.dx

where A = Cross sectional Area = πr² = π(18)² = 1017.88 ft²

dV = 1017.88 dx

The elemental force on that part will be

dF = ρg dV

ρg = 63.8 lbs/ft³

dF = 63.8 × 1017.88 dx = 64940.5 dx

F = ∫dF = 64940.5 dx

W = Fx = (∫dF)x = ∫ 64940.5x dx = 64940.5 ∫ xdx

We'll be integrating from (11 - 6) ft to 11 ft because that's the total height it'll be pumped through

W = 64940.5 (x²/2)¹¹₅ = 64940.5((11² - 5²)/2) = 3.12 × 10⁶ lbs.ft