Q85P - University Physics with Modern Physics

Question

A pump is required to lift 800 kg of water (about 210 gallons) per minute from a well 14.0 m deep and eject it with a speed of 18.0 m/s. (a) How much work is done per minute in lifting the water? (b) How much work is done in giving the water the kinetic energy it has when ejected? (c) What must be the power output of the pump?

Step-by-Step Solution

Answer

1. The work done is $1.10\times {10}^5\hspace{0.33em}J.$
2. The work done is $1.30\times {10}^5\hspace{0.33em}J$.
3. The power is 4.0 kW.

Step 1: Identify the given data

The mass is m = 800 kg.

The height is h = 14 m.

The velocity is v = 18 m/s.

Step 2: Concept/Significance of Work

When the external force is applied to an object then it gets moved by a distance in which transfer energy takes place. This is called work.

Step 3: Determine the work done per minute in lifting the water (a)

The expression to calculate the work done is given by,

 

W = mgh……………..(1)

 

Substitute 800 kg for m, $9.8 \mathrm{m}/{\mathrm{s}}^2$ for g, and 14 m for h in equation (1), we get,

 

$$\begin{gathered} W=\left(800 kg\right)\left(9.8 m/s^2\right)\left(14 m\right) \\ =109760 J \\ =1.10\times {10}^5 J \end{gathered}$$ 

 

Therefore, the required work done is $1.10\times {10}^5 J$.

Step 4: Determine the work done due to kinetic energy (b)

Find the kinetic energy the pump has when it has ejected the water as follows:

$KE=\frac{1}{2}m{v}^2$………………..(2)

Substitute 800 kg for m, and 18 m/s for v in equation (2).

$$\begin{gathered} \mathrm{W}=\frac{1}{2}\left(800 \mathrm{kg}\right){\left(18 \mathrm{m}\right)}^2 \\ =129600 \mathrm{J} \\ =1.30\times {10}^5 \mathrm{J} \end{gathered}$$ 

 

Therefore, the required work done is $1.30\times {10}^5 \mathrm{J}$.

Step 5: Determine the power output of the pump (c)

The power can be calculated as follows.

$$\begin{gathered} P=\frac{W_{total}}{t} \\ =\frac{\left(KE+W\right)}{t} \end{gathered}$$…………………..(3)

 

Substitute  $1.10\times {10}^5 J$ for W, $1.3\times {10}^5 J$ for KE, and 1min for t in equation (3), we get,

 

$$\begin{gathered} \mathrm{P}=\frac{\left(1.10\times {10}^5 \mathrm{J}+1.30\times {10}^5 \mathrm{J}\right)}{1 \min \left({\displaystyle \frac{60 \mathrm{s}}{1 \min }}\right)} \\ =4000 \mathrm{W}\left(\frac{1\mathrm{kW}}{1000 \mathrm{W}}\right) \\ =4.0 \mathrm{kW} \end{gathered}$$ 

 

Therefore, the required power is 4.0 kW.