Answer:
The work required to pump all of the water over the side is 2.0 x 10⁷ J
Step-by-step explanation:
Given;
diameter of the circular pool, D = 18 m
radius of the pool, r = 18/2 = 9m
depth of the pool, h = 4 m
depth of the side, H = 4 m
Work done = force x distance
Force of the water in the pool = ρVg
where;
ρ is the density of water = 1000 kg/m³
V is the volume of the pool
g is the acceleration due to gravity
Volume of the pool = πr²dy = π(9)²dy
Volume of the pool = 81π dy
Force = ρVg
Force = (1000) x (81πdy) x (9.8) = 793800π dy
Let the distance in which the water is moved = y
Work done = Fd = Fy
Work done = (793800π dy) x (y)
Work done = 793800πy dy
![Work\ done = \int\limits^4_0 {793800\pi y} \, dy\\\\Work\ done = 793800\pi \int\limits^4_0 {y} \, dy \\\\Work \ done = 793800\pi [(y^2)/(2)]^4_0\\\\Work \ done = (793800\pi)/(2) [y^2]^4_0\\\\Work \ done = 396900\pi (4^2-0)\\\\Work \ done = 396900\pi *16\\\\Work \ done = 6350400\pi \ J\\\\Work \ done = 19952956.8 \ J](https://img.qammunity.org/2021/formulas/physics/college/o8ydm47ve86cl4sexm7l5p158j9jqd4qm6.png)
Work done = 2.0 x 10⁷ J
Therefore, the work required to pump all of the water over the side is 2.0 x 10⁷ J