Question 6
A car with a mass of 1620 kg brakes so hard that its tyres slide on a horizontal road. The coefficient of kinetic friction between the tyres and the road is 0.75.
What is the kinetic frictional force between the tyres and the road, rounded to one decimal place?
Not quite. $0.75\times1620=1215\text{ N}$ uses the mass as if it were a force. The normal force is the weight, $mg$, so you need the 9.8. In 2025, 9.9% of Queensland students chose (A).
Not quite. $1620\div0.75=2160\text{ N}$ skips $g$ and divides by the coefficient. Friction is $\mu$ times the normal force, never the normal force divided by $\mu$. In 2025, 3.8% of Queensland students chose (B).
That is it. Step through the hints below to see why. In 2025, 82.8% of Queensland students chose (C).
Not quite. You found the normal force, 15 876 N, then divided by 0.75. A coefficient of friction below 1 always gives a friction force smaller than the normal force. In 2025, 3.1% of Queensland students chose (D).
The road is horizontal and nothing pushes the car up or down, so the road pushes back with the full weight.
$$F_N=mg=1620\times9.8=15\,876\text{ N}$$
The 1620 is a mass in kilograms. It only becomes a force once you multiply by $g$. That rules out (A) and (B).
The tyres are sliding, so this is kinetic friction. The formula book gives it directly.
$$F_f=\mu_k F_N=0.75\times15\,876=11\,907\text{ N}$$
With $\mu_k$ below 1, friction must come out smaller than the normal force. That rules out (D).
The options are in kilonewtons to one decimal place.
$$11\,907\text{ N}=11.907\text{ kN}\approx11.9\text{ kN}$$
Leave the rounding until this last line.
$$\begin{aligned}F_f&=\mu_k mg\\&=0.75\times1620\times9.8\\&=11\,907\text{ N}\approx11.9\text{ kN}\end{aligned}$$
On a horizontal road the normal force is the weight. Multiply it by the coefficient of kinetic friction and convert to kilonewtons.
The most common wrong answer was (A), from leaving out $g$. If a question gives you a mass and asks for a force, the 9.8 has to appear somewhere.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording and any figures are from the 2025 QCAA Engineering external assessment, © State of Queensland (QCAA) 2025, licensed under CC BY 4.0, and have been adapted, except where a third-party source is named. Tangent Tuition is not affiliated with the QCAA.
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