Question

point) A circular swimming pool has a diameter of 12 m. The circular side of the pool is 3 m high, and the depth of the water is 2.5 m. (The acceleration due to gravity is 9.8 m/s2m/s2 and the density of water is 1000 kg/m3kg/m3.) How much work (in Joules) is required to: (a) pump all of the water over the side? equation editorEquation Editor (b) pump all of the water out of an outlet 2 m over the side?

Answer 1

Answer:

(a) 86.65 J

(b) 149.65 J

Solution:

As per the question:

Diameter of the pool, d = 12 m

⇒ Radius of the pool, r = 6 m

Height of the pool, H = 3 m

Depth of the pool, D = 2.5 m

Density of water, $\rho_{w} = 1000\ kg//m^{3}$

Acceleration due to gravity, g = $9.8\ m/s^{2}$

Now,

(a) Work done in pumping all the water:

Average height of the pool, h = $\frac{H + D}{2}$

h = $\frac{3 + 2.5}{2} = 2.75\ m$

Volume of water in the pool, V = $\pi r^{2}h = \pi \times 6^{2}\times 2.75 = 311.02\ m^{3}$

Mass of water, $m_{w} = \frac{\rho_{w}}{V}$

$m_{w} = \frac{1000}{311.02} = 3.215\ kg$

Work done is given by the potential energy of the water as:

$W = m_{w}gh = 3.215\times 9.8\times 2.75 = 86.65\ J$

(b) Work done to pump all the water through an outlet of 2 m:

Now,

Height, h = 2.75 + 2 = 4.75

Work done,$W = m_{w}gh = 3.215\times 9.8\times 4.75 = 149.65\ J$