Question 19
An empty cable car with a mass of 1500 kg is being lowered down a track on a 25° incline at a constant velocity of 2 m/s when the cable snaps, releasing the cable car. The cable car’s emergency braking system engages when its velocity reaches 5 m/s and increases the coefficient of friction between the cable car and the track from 0.01 to 0.6, bringing it to a complete stop. A warning siren activates when the cable car is 10 metres from the original point of cable failure if it has not come to a stop.
Determine the stopped position of the cable car relative to the warning siren’s activation point. Answer to the nearest whole unit in metres.
[9 marks]The car speeds up for 2.59 m, brakes for 10.53 m, and stops 3 m past the siren
The track is marked every metre from the point where the cable snaps. Watch the orange friction arrow: tiny before the brakes, then bigger than the pull down the slope. The motion runs at half speed and the forces along the track are drawn to scale.
The velocity at the end of stage 1, 5 m/s, is the only thing that carries into stage 2. The forces are worked out again from scratch with the new coefficient.
Tune the brakes and see where the car stops
Set the speed at which the brakes engage, the braked coefficient, the slope and the mass. The siren stays at 10 m. The question is 5 m/s, 0.6, 25° and 1500 kg.
This is the question. The car covers 2.59 m speeding up and 10.53 m braking, 13 m in all, so it stops 3 m past the siren point.
Every force changed with the mass, but both accelerations stayed the same, so the car still stops in the same place. Mass cancels in $a=F_{net}\div m$ on a slope with only friction acting.
The brakes cannot stop the car. Braked friction is no bigger than the pull down the slope, so the net force still points down the track. To stop on this slope the coefficient has to be more than $\tan\theta$.
Braking later means a longer stage 1 and a faster start to stage 2, and the stopping distance goes with the square of that speed. A grippier brake shortens stage 2. A steeper slope weakens the brake because friction shrinks while the pull grows.
Work through the solution one mark at a time
Try each step yourself before you reveal it. Stage 1 runs from the cable failure to the brakes, stage 2 from the brakes to the stop.
Once the cable snaps, the only forces along the track are friction and the pull of gravity. Start with the normal force, then the friction before the brakes.
$$\begin{aligned}F_N&=mg\cos\theta\\&=1500\times9.8\times\cos25^\circ\\&=13\,322.72\text{ N}\end{aligned}$$
$$\begin{aligned}F_f&=\mu F_N\\&=0.01\times13\,322.72\\&=133.23\text{ N}\end{aligned}$$
The normal force is the same in both stages. Only the coefficient changes.
Marker: correctly determines the initial frictional force. QCAA’s note: “Accept initial frictional force range from 133.2272447 to 133.23 N.”
$$\begin{aligned}F_\parallel&=mg\sin\theta\\&=1500\times9.8\times\sin25^\circ\\&=6212.49\text{ N}\end{aligned}$$
QCAA calls this $F_P$, the force down the incline.
Marker: correctly determines the force down the incline. QCAA’s note: “Accept force down the incline range from 6212.488448 to 6212.49 N.”
The car slides down, so friction acts up the track against it.
$$\begin{aligned}F_{net}&=6212.49-133.23\\&=6079.26\text{ N}\end{aligned}$$
Marker: correctly determines the net force. QCAA’s note: “Accept net force range from 6079.26 to 6079.261203 N.”
$$a=\frac{F_{net}}{m}=\frac{6079.26}{1500}=4.05\text{ m s}^{-2}$$
QCAA’s sample rounds this to 4.1, one decimal place where the rest of its working keeps two. The guide accepts anything from 4.05 to 4.1. Keep 4.05 or better.
Marker: correctly determines the acceleration. QCAA’s note: “Accept acceleration range from 4.05 to 4.1 m/s².”
The car was already moving at 2 m/s when the cable snapped, so $u=2$, not 0. It reaches $v=5$ when the brakes engage.
$$s_1=\frac{v^2-u^2}{2a}=\frac{5^2-2^2}{2\times4.05}=2.59\text{ m}$$
QCAA’s 4.1 gives 2.56 m. Both are in the accepted range of 2.56 to 2.6 m.
Marker: correctly determines the displacement before brakes engage. QCAA’s note: “Accept displacement range from 2.56 to 2.6 m.”
New coefficient, same normal force. The pull down the slope has not changed either.
$$F_f=0.6\times13\,322.72=7993.63\text{ N}$$
$$\begin{aligned}F_{net}&=6212.49-7993.63\\&=-1781.14\text{ N}\end{aligned}$$
Negative means the net force now points up the track, so the car slows down.
Marker: correctly determines the net force after brakes engage. QCAA’s note: “Accept net force after brakes engage range from -1781.14 to -1781.146234 N.”
$$a=\frac{-1781.14}{1500}=-1.19\text{ m s}^{-2}$$
Marker: correctly determines the deceleration after brakes engage. QCAA’s notes: “Accept acceleration after brakes engage range from -1.187430823 to -1.19 m/s²” and “Allow FT error for incorrect net force after braking when determining acceleration.”
Carry the 5 m/s across as the new $u$. The car comes to a complete stop, so $v=0$.
$$\begin{aligned}s_2&=\frac{v^2-u^2}{2a}\\&=\frac{0^2-5^2}{2\times-1.19}\\&=10.50\text{ m}\end{aligned}$$
With the unrounded deceleration it is 10.53 m. The guide accepts 10.50 to 10.53 m.
Marker: correctly determines the displacement after brakes engage. QCAA’s note: “Accept displacement after brakes engage range from 10.50 to 10.52692903 m.”
Add the two stages, then compare with the siren point at 10 m.
$$\begin{aligned}s_1+s_2&=2.59+10.50\\&=13.09\text{ m}\approx13\text{ m}\end{aligned}$$
QCAA’s sample adds $2.56+10.50=13.06$ m and full precision gives $2.59+10.53=13.12$ m. Every route lands on 13 m. The question asks for the position relative to the siren, so 13 m on its own is not the answer. Say which side of the siren it stops on.
Marker: correctly determines the answer to the nearest whole unit. QCAA’s note: “Allow FT error for incorrect … displacement when determining the answer.”
Each phrase of the question gave you something
QCAA did not publish a band for this question. It reads as complex unfamiliar because the stem never tells you there are two stages. You have to split the slide at the brakes, run a full force-to-motion chain twice with a velocity handed between them, and then turn the total into a position relative to a point the question only mentions in passing.
If I had to guess where you might lose marks, it would be the start of stage 1. The car is already doing 2 m/s when the cable snaps, and starting from rest gives a stage 1 of 3.08 m and an answer of 4 m past the siren. The other trap is the last line. Writing 13 m answers a different question, and forgetting stage 1 altogether puts the car only 1 m past the siren.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording, any figures and marking-guide steps are from the 2022 QCAA Engineering external assessment, © State of Queensland (QCAA) 2022, 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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