To find the work required to pump all of the water out of the pool, we need to calculate the total work done in moving the water to the top of the pool. Work is calculated as the force times the distance.

Let's break this problem down step-by-step:

Step 1: Determine the volume of the water in the pool.

The pool is a cylinder, so the volume $V$ of the water can be calculated using the formula for the volume of a cylinder:

$V = \pi r^2 h$

where:

• $r$ is the radius of the pool,
• $h$ is the height of the water.

Given:

• Diameter of the pool = 20 ft, so the radius $r = \frac{20}{2} = 10 \, \text{ft}$,
• Depth of the water = 4 ft (this is the height $h$ of the water).

So, the volume of water is:

$V = \pi (10)^2 (4) = 400\pi \, \text{ft}^3$

Now, the weight of the water is given by:

$\text{Weight} = \text{Volume} \times \text{Weight per unit volume}$

The weight per unit volume of water is 62.5 lb/ft³, so:

$\text{Weight} = 400\pi \times 62.5 = 25,000\pi \, \text{lb}$

Step 2: Work calculation.

Work is calculated using the formula:

$\text{Work} = \int_0^h F(x) \, dx$

where:

• $F(x)$ is the force required to lift a small volume of water at height $x$ from the bottom of the pool to the top.

The small volume of water at height $x$ is a thin slice of the pool, which has a volume $\Delta V = \pi r^2 \, \Delta x$. The force to lift this small volume is the weight of this slice, which is given by:

$\text{Force at height } x = \pi r^2 \, \Delta x \times 62.5$

This slice must be lifted a distance of $(5 - x)$ feet to the top of the pool (since the height of the pool is 5 feet).

Thus, the work to lift the slice of water is:

$\text{Work at height } x = \pi r^2 \, 62.5 \, (5 - x) \, dx$

Now, we can integrate from $x = 0$ to $x = 4$ (the depth of the water):

$\text{Work} = \int_0^4 \pi (10)^2 62.5 (5 - x) \, dx$

Simplifying the constants:

$\text{Work} = 6250\pi \int_0^4 (5 - x) \, dx$

Now, evaluate the integral:

$\int_0^4 (5 - x) \, dx = \left[ 5x - \frac{x^2}{2} \right]_0^4$

$= \left( 5(4) - \frac{4^2}{2} \right) - (0) = 20 - 8 = 12$

So, the total work is:

$\text{Work} = 6250\pi \times 12 = 75,000\pi \, \text{ft-lb}$

Approximating $\pi \approx 3.1416$:

$\text{Work} \approx 75,000 \times 3.1416 = 235,620 \, \text{ft-lb}$

Final Answer:

The work required to pump all the water out is approximately 235,620 ft-lb.

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Would you like further clarification or details on any step? Here are a few related questions:

1. How do the units of force and work relate in this context?
2. What would happen if the depth of the water was 5 ft instead of 4 ft?
3. How can this method be applied to calculate the work for a non-circular pool?
4. What if the weight of the water were different, say 62 lb/ft³ instead of 62.5 lb/ft³?
5. How would the total work change if the water were being pumped to a higher or lower height?

Tip: When solving for work, always remember to account for how the force varies with height, especially in fluids where the depth influences the volume being lifted!