91Ó°ÊÓ

• Speed of water through the hose is $\mathrm{v}=5.0\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$.

• Radius of the hose is $\mathrm{r}=0.1\mathrm{cm}=0.01\mathrm{m}$.

• The height of the window above waterline is $\mathrm{h}=3.0\mathrm{m}$.

Use the work-energy theorem, equation of power relating to work and time, and the mass flow rate of water to find the power of the pump. According to the work-energy theorem, the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.

Formulae are as follows:

$\mathrm{Δ°Â}=\mathrm{Δ°­}+\mathrm{Δ\pm «}$

$\mathrm{Δ°­}=\frac{1}{2}{\mathrm{mv}}^2$

$\mathrm{Δ\pm «}=\mathrm{mgh}$

$\mathrm{P}=\frac{\mathrm{Δ°Â}}{\mathrm{Δ³Ù}}$

$\frac{\mathrm{Δ³¾}}{\mathrm{Δ³Ù}}−\mathrm{ÒÏ´¡\pm ¹}$

Where the is mass, is velocity, is area, is time, is density, is the kinetic energy, is the potential energy, is the work done, is an acceleration due to gravity, is height, and is the power.

From the work-energy theorem,

$\mathrm{Δ°Â}=\mathrm{Δ°­}+\mathrm{Δ\pm «}$

And, the equation of power is,

$\mathrm{P}=\frac{\mathrm{Δ°Â}}{\mathrm{Δ³Ù}}$

$=\frac{\mathrm{Δ°­}+\mathrm{Δ\pm «}}{\mathrm{Δ³Ù}}$

$=\frac{\mathrm{Δ³¾}}{\mathrm{Δ³Ù}}\left(\mathrm{gh}+\frac{1}{2}{\mathrm{v}}^2\right)$

Now, the rate of mass flow of fluid is given by,

$\frac{\mathrm{Δ³¾}}{\mathrm{Δ³Ù}}−\mathrm{ÒÏ´¡\pm ¹}$

So, the equation of power will become,

$\mathrm{gh}\left(\mathrm{v}+\frac{1}{2}2\right)$

$=1000×\mathrm{π}×(0.01\mathrm{m}{)}^2×(5\mathrm{m}/\mathrm{s})\left(9.8\mathrm{m}/{\mathrm{s}}^2×3.0\mathrm{m}+\frac{1}{2}(5.0\mathrm{m}/\mathrm{s}{)}^2\right)$

$=65.8\mathrm{W}$

Hence, power of the pump is, $65.8\mathrm{W}$.