Question

A circular swimming pool has a diameter of 18 m, the sides are 4 m high, and the depth of the water is 4 m. How much work (in Joules) is required to pump all of the water over the side

Answer 1

The work required to pump all of the water over the side is 3.991×10^7 Joules.

Step-by-step explanation:

Data given:

Diameter (D) = 18 m

Height = 4 m

Depth = 4 m

Work required to pump all of the water over the side is calculated by multiplying the pressure in the pool by the volume of the pool

Pressure (P) = height of swimming pool × density of water × acceleration due to gravity = 4 m × 1000 kg/m^3 × 9.8 m/s^2 = 39,200 kg/m.s^2 = 39,200 N/m^2

Volume (V) = area of pool × depth = πD^2/4 × depth = 3.142×18^2/4 × 4 = 1018.008 m^3

Work required = PV = 39,200 N/m^2 × 1018.008 m^3 = 3.991×10^7 Nm = 3.991×10^7 J

Answer 2

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`

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