Mathematics 9709/41 — May/June 2014

Walkthrough

This problem asks for the speed of a train moving at constant speed, given the engine power and the total resistance.

The first step is to make sure all quantities are in consistent SI units. Power is given in kilowatts, so convert it to watts by multiplying by 1000. Resistance is given in kilonewtons, so convert it to newtons, also by multiplying by 1000.

Next, recall that when a vehicle moves at constant speed, its acceleration is zero. Newton's second law then tells us that the resultant force on the train is zero. Therefore, the driving force produced by the engine must exactly balance the total resistance. This gives the value of the driving force $DF$ directly as $28\,000\text{ N}$.

The key relationship is power = force $\times$ speed, i.e. $P = DF \times V$. Substituting the known values gives a simple linear equation for $V$.

Finally, divide both sides by $28\,000$ to isolate $V$. The calculation gives $47.5$, so the train travels at $47.5\text{ m s}^{-1}$.

Key Takeaways

The main skill is applying the power formula $P = Fv$ in a mechanics context. It is also important to remember that constant speed means zero acceleration, so the driving force equals the total resistance. Unit conversion is essential: power must be in watts and force in newtons before using the formula.

Common Mistakes

• Forgetting to convert kilowatts to watts, which would give a completely wrong value for $V$.
• Forgetting to convert kilonewtons to newtons, leading to an incorrect driving force.
• Writing the power formula incorrectly, for example using $P = \frac{F}{v}$ or $P = Fv^2$.
• Assuming the train must accelerate, and trying to use Newton's second law with acceleration instead of recognising the constant-speed condition.

Things to Be Careful About

The units must be consistent: use watts, newtons and metres per second. The value $1330\text{ kW}$ becomes $1\,330\,000\text{ W}$, and $28\text{ kN}$ becomes $28\,000\text{ N}$. Also note that the mark scheme expects the driving force to be stated as $28\,000\text{ N}$ before applying $P = DF \times V$. The final speed is exactly $47.5\text{ m s}^{-1}$; an unsupported answer may not receive full credit, so show the equation used.