CALCULATE WORK SOLVE IN PAPER CLEARLY

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 19.5 foot radius. It is 12 feet tall and has 7 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=63.6𝑙𝑏𝑠/𝑓𝑡3σ=63.6lbs/ft3 Don't forget to enter the correct units. (You may enter lbf or lb*ft for ft-lb.)

Answer 1

It is unfortunate that the problem statement uses imperial unit, rather than metric, and

it is even more unfortunate that it uses "lbs" ambiguously --

sometimes it is a unit of mass, and sometimes a unit of force.

I will use "lb" to be a unit of mass.

Let

r = 19.5 ft be the radius of the pool,

σ = 63.6 lb/ft³ be the water density,

g = 32 ft/s² be gravitational acceleration.

As the water is being pumped out, its level starts at 5 ft below the side of the pool,

and ends 12 ft below side of the pool.

Consider the situation when the water is at some depth h below the side of the pool.

And imagine that from the water surface you are going to pump out a very thing layer of thickness Δh.

The volume V of that layer is

V = πr²Δh

The mass m of that layer is

m = Vσ = πr²σΔh

That mass needs to be lifted the height h.

From physics the work required to lift mass m distance h is

W(h) = mgh = πr²σghΔh

To calculate the work needed for the whole pool, we sum up the work needed for

all the thin layer, and

take the limit of the sum as Δh -> 0.

That is the definite integral from 5 to 12 of

F(h) = ∫W(h)dh = ∫πr²σghdh = πr²σgh²/2

The definite integral is

F(12) - F(5) = πr²σg((12ft)² - (5ft)²)/2

Substituting actual numbers, the total work is

π×(19.5ft)²×(63.6lb/ft³)×(32ft/s²)×((12ft)² - (5ft)²)/2 = 144658232 lb*ft²/s²