Chapter 10.1, Problem 12PP

To determine

The power developed by a pimp in lifting $35\text{L}$ of water per minute from a depth of $110\text{m}$ .

Answer to Problem 12PP

  $628.834\text{W}$

Explanation of Solution

Given:

The mass of water being lifted per minute, $m=35\text{L = 35}\left(1.00\text{kg}\right)=35\text{kg}$

Depth through which it is lifted, $d=110\text{m}$

Formula used:

Power is given by the work done divided by the taken to do the work,

  $P=\frac{W}{t}$

Where $W$ is the work done, and $t$ is the time taken

Also, work done is given by,

  $W=Fd\cos \theta$

Where $F$ is the force done, $d$ is the distance, and $\theta$ is the angle between the distance and force.

Calculation:

Now, when the water is being lifted then the force acting on it is due to its weight,

  $F=mg$

Then, work done will be,

  $\begin{array}{c}W=Fd\cos \theta \\ =mgd\cos 0°\\ =mgd\end{array}$

Then, power needs to be developed by the pump is,

  $\begin{array}{c}P=\frac{W}{t}\\ =\frac{mgd}{1\min }\\ =\frac{\left(35\right)\left(9.8\right)\left(110\right)}{60\text{s}}\\ =628.834\text{W}\end{array}$

Conclusion:

Thus, the power that needs to be developed by the pump is $628.834\text{W}$ .