(II) A pump lifts 27.0 kg of water per minute through a height of 3.50 m. What minimum output rating (watts) must the pump motor have?

In this problem, the amount of force needed to raise the water is equal to the weight of the water. Also, the value of time is one minute in this case.

Given data:

The mass of water is\(m = 27\;{\rm{kg}}\).

The height is\(d = 3.50\;{\rm{m}}\).

The relation of work done in raising the water is given by:

\(W = mgd\)

Here, g is the gravitational acceleration.

The relation for findingthe minimum output rating is given by:

\(P = \frac{W}{t}\)

On plugging the values in the above relation, you get:

\(\begin{aligned}P &= \left( {\frac{{mgd}}{t}} \right)\\P &= \left( {\frac{{\left( {27\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {3.50\;{\rm{m}}} \right)}}{{1\;{\rm{min}} \times \frac{{60\;{\rm{s}}}}{{1\;{\rm{min}}}}}}} \right)\\P &= 15.4\;{\rm{W}}\end{aligned}\)

Thus, \(P = 15.4\;{\rm{W}}\) is the minimum output rating of the motor.