Solved Problem on Work and Energy

To draw water from a well, a pump with a power of 3675 W is used. The well depth is 30 m. Calculate the volume of water taken in 24 h. The density of water is 1000 kg/m³.

Problem data:

Power of pump: \( \mathscr{P} \) = 3675 W;
Depth of well: h = 30 m;
Pump operating time interval: Δ t = 24 h;
Density of water: ρ = 1000 kg/m³;
Acceleration due to gravity: g = 9.8 m/s².

Problem diagram:

We choose a Reference Level (RL) on the surface of the well (Figure 1).

Figure 1

Solution:

First, we convert the interval of time given in hours (h) to seconds used in the International System of Units (SI).

\[ \begin{gather} \Delta t=24\;\mathrm{\cancel h}\times\;\frac{60\;\mathrm{\cancel{min}}}{1\;\mathrm{\cancel h}}\times\frac{60\;\mathrm s}{1\;\mathrm{\cancel{min}}}=86400\;\mathrm s \end{gather} \]

Power is given by energy variation per unit of time.

\[ \begin{gather} \bbox[#99CCFF,10px] {\mathscr{P} =\frac{\Delta E}{\Delta t}} \tag{I} \end{gather} \]

The change of total energy will be given by the change of potential energy.

\[ \begin{gather} \bbox[#99CCFF,10px] {\Delta E=U_f-U_i} \tag{II} \end{gather} \]

the potential energy is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {U=mgh} \tag{III} \end{gather} \]

substituting equation (III) into equation (II), the change in potential energy when a mass of water is taken from the bottom of the well (reference level, h₀ = 0) until the surface will be

\[ \begin{gather} \Delta E=mgh-mgh_0\\[5pt] \Delta E=mgh-mg\times 0\\[5pt] \Delta E=mgh \tag{IV} \end{gather} \]

substituting the equation (IV) into equation (I)

\[ \begin{gather} \mathscr{P} =\frac{mgh}{\Delta t}\\[5pt] m=\frac{\mathscr{P} \Delta t}{gh}\\[5pt] m=\frac{(3675\;\mathrm W)\times(86400\;\mathrm s)}{\left(9.8\;\frac{\mathrm m}{\mathrm{s^2}}\right)\times(30\;\mathrm m)}\\[5pt] m=1080000\;\text{kg} \end{gather} \]

The density is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {\rho =\frac{m}{V}} \end{gather} \]

The volume of water removed will be

\[ \begin{gather} V=\frac{m}{\mu}\\[5pt] V=\frac{1080000\;\mathrm{\cancel{kg}}}{1000\;\mathrm{\frac{\cancel{{kg}}}{m^3}}} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {V=1080\;\mathrm m^3} \end{gather} \]

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .