Mathematics 9709/42 — October/November 2022

Walkthrough

Start by identifying what energy is involved. The road $AB$ is horizontal, so the height of the cyclist does not change and there is no change in gravitational potential energy. The only energy change is the increase in kinetic energy, which starts at $0$ because the cyclist begins from rest at $A$.

First, find the work done by the driving force. The force of $100\text{ N}$ acts in the same direction as the motion, so the angle between the force and the displacement is $0°$ and $W = Fd\cos 0° = Fd = 100 \times 50 = 5000\text{ J}$. This is the total energy supplied to the system by the cyclist.

Not all of this energy becomes kinetic energy: some of it is spent overcoming the resistance. The question gives the work done against the resistance as $3560\text{ J}$. The net work available to increase speed is therefore:

$$5000 - 3560 = 1440\text{ J}$$

The work–energy principle states that the net work done on an object equals its change in kinetic energy:

$$1440 = \frac{1}{2}mv^2$$

With $v = 6$:

$$\frac{1}{2}m(36) = 1440$$ $$18m = 1440$$ $$m = 80\text{ kg}$$

As a cross-check, the alternative SUVAT approach gives the same result. Assuming constant acceleration, $v^2 = u^2 + 2as$ with $u = 0$, $v = 6$, $s = 50$ yields:

$$a = \frac{36}{100} = 0.36\text{ m s}^{-2}$$

The resistance force is $\frac{3560}{50} = 71.2\text{ N}$. Newton's second law, $100 - 71.2 = 0.36m$, gives $m = \frac{28.8}{0.36} = 80\text{ kg}$, confirming the answer.

Key Takeaways

• The work–energy principle: net work done on an object equals its change in kinetic energy.
• Work done by a constant force: $W = Fd\cos\theta$; when the force is along the motion, $W = Fd$.
• On a horizontal road there is no change in gravitational potential energy, so all net work becomes kinetic energy.
• Units used: work and energy in joules (J), mass in kilograms (kg), speed in $\text{m s}^{-1}$.

Common Mistakes

• Omission of the resistance term: writing $5000 = \frac{1}{2}mv^2$ ignores the $3560\text{ J}$ and produces the wrong mass.
• Sign errors in the energy equation. The mark scheme allows sign errors for the method mark but still needs the three-term equation to be dimensionally correct.
• Confusing 'work done by the resistance' with 'work done against the resistance'.
• In the SUVAT alternative, applying Newton's second law without including both the driving force and the resistance.

Things to Be Careful About

• The force and displacement are both along the road, so $\cos\theta = 1$ and $W = Fd$ directly.
• The speed is squared: the kinetic energy is $\frac{1}{2}m \times 6^2$, not $\frac{1}{2} m \times 6$.
• Keep all units consistent; the mass must be expressed in kilograms.
• Show the work done by the cyclist ($5000\text{ J}$) explicitly, since the mark scheme awards the first mark for it.