To solve this problem it is necessary to apply the concepts related to the flow as a function of the volume in a certain time, as well as the potential and kinetic energy that act on the pump and the fluid.
The work done would be defined as
[tex]\Delta W = \Delta PE + \Delta KE[/tex]
Where,
PE = Potential Energy
KE = Kinetic Energy
[tex]\Delta W = (\Delta m)gh+\frac{1}{2}(\Delta m)v^2[/tex]
Where,
m = Mass
g = Gravitational energy
h = Height
v = Velocity
Considering power as the change of energy as a function of time we will then have to
[tex]P = \frac{\Delta W}{\Delta t}[/tex]
[tex]P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)[/tex]
The rate of mass flow is,
[tex]\frac{\Delta m}{\Delta t} = \rho_w Av[/tex]
Where,
[tex]\rho_w[/tex] = Density of water
A = Area of the hose [tex]\rightarrow A=\pi r^2[/tex]
The given radius is 0.83cm or [tex]0.83 * 10^{-2}[/tex]m, so the Area would be
[tex]A = \pi (0.83*10^{-2})^2[/tex]
[tex]A = 0.0002164m^2[/tex]
We have then that,
[tex]\frac{\Delta m}{\Delta t} = \rho_w Av[/tex]
[tex]\frac{\Delta m}{\Delta t} = (1000)(0.0002164)(5.4)[/tex]
[tex]\frac{\Delta m}{\Delta t} = 1.16856kg/s[/tex]
Final the power of the pump would be,
[tex]P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)[/tex]
[tex]P = (1.16856)((9.8)(3.5)+\frac{1}{2}5.4^2)[/tex]
[tex]P = 57.1192W[/tex]
Therefore the power of the pump is 57.11W
