# Biophysics Problem 28

A donkey is used to raise water from a well. He is walking around a circle of \(2 \;m\) radius at a speed of \(3\; km /hr.\) He is raising \(100 \;kg\) of water a distance of \(2 \;m\) every second. What force is he exerting on the lever arm?

This problem is not really as difficult as it looks. The work done by the donkey in one second will equal the increase the potential energy of the water in the same time interval.

Calculate the increase of potential energy of the water in one second.

The increase in potential energy of the water \((PE)\) in one second will be:

\(PE = m\; g\; h \\ PE = (100 \;kg) (9.8\; m/s²) (2\;m) \\ PE = 1960\)

The work done by the donkey will equal the force he exerts multiplied by the distance he travels in one second.

How far does the donkey travel in one second?

If the donkey travels \(3000 \;m\) in one hour. then in one second, the donkey travels:

\(\frac{3000\;m}{h} \times \frac{1\; h}{3600\;s}= 0.833\;m/s\)

Now calculate the donkey's force in Newtons.

Remember, the work done by the donkey equals the increase of potential energy of the water in the same one-second time interval. Therefore,

\(F \;s = m\; g \;h \\ F = (m\; g \;h) / s \\ F = (1960 \;J) / (0.833\; m) \\ F = 2350 \;N\)