Solved Problem on Work and Energy
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To draw water from a well, it is used a pump with a power of 3675 W. 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
\[ \begin{gather} \bbox[#99CCFF,10px] {\rho =\frac{m}{V}} \end{gather} \]
\[ \begin{gather} V=\frac{m}{\mu}\\[5pt] V=\frac{1080000\;\mathrm{\cancel{kg}}}{1000\;\mathrm{\frac{\cancel{{kg}}}{m^3}}} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {V=1080\;\mathrm m^3} \end{gather} \]
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .