Complete Question
A circular swimming pool has a diameter of 16 m. The circular side of the pool is 4 m high, and the depth of the water is 3.5 m. (The acceleration due to gravity is 9.8 m/s2 and the density of water is 1000 kg/m3.) How much work (in Joules) is required to:
(a) pump all of the water over the side
(b) pump all of the water out of an outlet 3 m over the side?
Answer:
a
The work required is 
b
The work is
Explanation:
From the question we are told that
The diameter is
The height of the circular side is 
The depth of water is 
The acceleration due to gravity is 
The density of water is
The radius of the swimming pool is

The work is mathematically given as

Now force is mathematically given as


Now the work done to pump all of the water over the side
is



![= 64000 * 9.8 \pi [4x - \frac{x^2}{2} ]\left {3.5} \atop {0} \right.](https://tex.z-dn.net/?f=%3D%2064000%20%2A%209.8%20%5Cpi%20%5B4x%20-%20%5Cfrac%7Bx%5E2%7D%7B2%7D%20%5D%5Cleft%20%7B3.5%7D%20%5Catop%20%7B0%7D%20%5Cright.)
![= 64000 *9.8 \pi [4(3.5) - \frac{3.5^2}{2} ]](https://tex.z-dn.net/?f=%3D%2064000%20%2A9.8%20%5Cpi%20%5B4%283.5%29%20-%20%5Cfrac%7B3.5%5E2%7D%7B2%7D%20%5D)

Now for 3 meters over the side the new height would be



Now applying the formula for work that is above


![= 64000 * 9.8 \pi [7x - \frac{x^2}{2} ]\left {3.5} \atop {0} \right.](https://tex.z-dn.net/?f=%3D%2064000%20%2A%209.8%20%5Cpi%20%5B7x%20-%20%5Cfrac%7Bx%5E2%7D%7B2%7D%20%5D%5Cleft%20%7B3.5%7D%20%5Catop%20%7B0%7D%20%5Cright.)
![= 64000 *9.8 \pi[7(3.5) - \frac{(3.5)^2}{2} ]](https://tex.z-dn.net/?f=%3D%2064000%20%2A9.8%20%20%5Cpi%5B7%283.5%29%20-%20%5Cfrac%7B%283.5%29%5E2%7D%7B2%7D%20%5D)
