A tractor engine develops 45 kW of power at the drawbar. If it pulls a plough that requires a draft force of 9 kN, what is the operating speed of the tractor?

Using the relation Power = Force × Velocity: $V = \frac{P}{F} = \frac{45{,}000 \text{ W}}{9{,}000 \text{ N}} = 5 \text{ m/s}$. The distractor 3 m/s results from dividing 45 by 9 incorrectly with wrong unit conversion, 9 m/s from confusing force with power, and 45 m/s from ignoring the division entirely.