Respuesta :
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 [tex]W_d = 14.25*10^6J[/tex]
b
The work is [tex]W_{d3} = 3.6 *10^7 J[/tex]
Explanation:
From the question we are told that
The diameter is [tex]d = 16m[/tex]
The height of the circular side is [tex]H =4m[/tex]
The depth of water is [tex]D = 3.5m[/tex]
The acceleration due to gravity is [tex]g =9.81m/s^2[/tex]
The density of water is [tex]\rho_w = 1000kg/m^3[/tex]
The radius of the swimming pool is
[tex]r = \frac{16}{2} = 8m[/tex]
The work is mathematically given as
[tex]W_d = Force * distance[/tex]
Now force is mathematically given as
[tex]F = density * area * height \ of \ pool[/tex]
[tex]= \rho * (\pi r^2 ) dx[/tex]
Now the work done to pump all of the water over the side
is
[tex]W_d = \int\limits^D_0 {\rho (\pi r^2 )(H-x)} \, dx[/tex]
[tex]=\int\limits^D_0 {1000*9.81 *(\pi * 8^2) (4- x)} \, dx[/tex]
[tex]= 64000 *9.8 \pi \int\limits^{3.5} _0 {(4-x)} \, dx[/tex]
[tex]= 64000 * 9.8 \pi [4x - \frac{x^2}{2} ]\left {3.5} \atop {0} \right.[/tex]
[tex]= 64000 *9.8 \pi [4(3.5) - \frac{3.5^2}{2} ][/tex]
[tex]=14.25*10^6J[/tex]
Now for 3 meters over the side the new height would be
[tex]H_{new} = H +3[/tex]
[tex]= 4+3[/tex]
[tex]=7m[/tex]
Now applying the formula for work that is above
[tex]W_{d3 }= \int\limits^D_0 {\rho g (\pi r^2)(H-x)} \, dx[/tex]
[tex]= \int\limits^{3.5}_0 {1000 *9.8 * (\pi * 8^2)(7 - x)} \, dx[/tex]
[tex]= 64000 * 9.8 \pi [7x - \frac{x^2}{2} ]\left {3.5} \atop {0} \right.[/tex]
[tex]= 64000 *9.8 \pi[7(3.5) - \frac{(3.5)^2}{2} ][/tex]
[tex]=3.6*10^7J[/tex]
