Work, Energy and Power: Application Problems (7)

Question

A pump lifts $200\,\text{kg}$ of water per minute from a depth of $10\,\text{m}$ and delivers it with a speed of $20\,\text{m/s}$. Find the power of the pump. Take $g = 10\,\text{m/s}^2$.

Solution — Step by Step

The power of the pump is $1000\,\text{W}$ or $1\,\text{kW}$.

Why This Works

Power is energy per unit time. Whenever you see “per minute” or “per second” in the question, switch your thinking to rates. The water gains both PE (lifted height) and KE (delivery speed), so total mechanical energy per second is $\dot{m}(gh + \tfrac{1}{2}v^2)$.

The mass per second is the natural quantity for flow problems. Once you have $\dot{m}$, multiplying by energy per kilogram gives power directly.

Alternative Method

Compute total energy per minute, then divide by $60$. Energy per minute $= 200 \times (100 + 200) = 60{,}000\,\text{J}$. Power $= 60000/60 = 1000\,\text{W}$. Same answer, slightly slower.

Common Mistake

Forgetting the kinetic energy term. Many students compute only $P = \dot{m}gh = 333\,\text{W}$ and pick the wrong option. The phrase “delivers it with a speed of $20\,\text{m/s}$” is the giveaway — if speed is mentioned, KE matters.