Question 20
A motor operating a platform lift is connected to a single-thread worm and a worm wheel with 60 teeth and a diameter of 150 mm, as shown. The worm wheel is directly attached to a rack to slowly raise and lower the platform lift. It takes 25 seconds to raise the platform lift to a height of 2 m.
The combined mass of the platform and the load it is lifting is 250 kg.
The motor uses 280 W of power to raise the combined mass to a height of 2 m.
Each turn of the worm wheel moves the rack one circumference, so 2 m takes 4.24 turns
The wheel’s 60 teeth, the rack and the worm thread are drawn at the same pitch, so they mesh. Watch the orange dot count the wheel’s turns while the gauge on the right climbs to 2 m.
The 60 teeth and the single thread set how fast the worm must spin, which no part asks for. Part a) needs only the wheel’s circumference.
Change the wheel, the load and the motor, then try the slips
The ruler splits the lift into one segment per wheel turn. The question is a 150 mm wheel, 2 m, 250 kg, 25 s and 280 W.
This is the question. 4.24 wheel turns raise the platform 2 m. The lift gains 4900 J in 25 s, which is 196 W of useful power from the motor’s 280 W, so the lift is 70% efficient.
A bigger wheel needs fewer turns for the same lift. A heavier load or a faster lift needs more useful power, so with the same motor power the efficiency rises. Past 100% the numbers describe a lift that cannot exist.
The 150 mm is a diameter. Putting it into $2\pi r$ doubles the circumference and halves the turns, which falls outside QCAA’s 4.2 to 4.3. The guide accepts $\pi d$ or $2\pi r$, so use 0.075 m in $2\pi r$.
Input over output flips the efficiency. At the question’s values it gives $280\div196=1.43$, or 143%, which would mean the lift makes energy. Efficiency is always the useful output over the input, so it must come out below 1.
Multiplying by 60 gives the turns of the worm, not the worm wheel. The single thread and the 60 teeth tell you how fast the worm spins, and part a) does not ask for that.
Work through the solution one mark at a time
Try each step yourself before you reveal it.
The wheel meshes with the rack, so each turn of the wheel moves the rack one circumference of the wheel. Write that as a formula.
$$\begin{aligned}&\text{turns of the worm wheel}\\&=\frac{\text{linear distance}}{\text{circumference of wheel}}\\&=\frac{2}{\pi d}\end{aligned}$$
The guide accepts $\pi d$ or $2\pi r$ for the circumference, and this mark can be implied by your working.
Marker: correct formula identified to calculate worm wheel rotations
Convert the diameter to metres, 150 mm = 0.15 m, then divide.
$$\frac{2}{0.15\pi}=4.244\text{ turns}$$
Keep the decimals. Rounded to a whole number, 4 turns falls outside the accepted range. The 60 teeth and the single thread are not needed here.
Marker: correctly calculates the number of rotations of the worm wheel to raise the platform 2 m vertically. “Accept 4.2–4.3 inclusive.”
Work done at the lift is the output side: the load force is the weight, $mg$, and it rises 2 m.
$$\begin{aligned}W&=F_L\,d_L=mgh\\&=250\times9.8\times2\end{aligned}$$
The guide also accepts $W=F_E\,d_E$, but its worked version assumes 100% efficiency so that it lands back on $mgh$. Part c) shows the lift is 70% efficient, so the work at the motor is bigger. “At the lift” means the output side, so use $mgh$.
Marker: calculates the work done at the lift to raise the combined mass to the maximum height of 2 m
The 280 W is a power, so turn the 4900 J into a power too, using the 25 s from the stem.
$$P=\frac{W}{t}=\frac{4900}{25}=196\text{ W}$$
Marker: correctly calculates the useful power output
Useful output over input, then times 100%.
$$\eta=\frac{P_{\text{out}}}{P_{\text{in}}}\times100\%=\frac{196}{280}\times100\%$$
The energy route gives the same answer: the motor supplies $280\times25=7000$ J, and $\tfrac{4900}{7000}=0.7$. The guide accepts either route, and it accepts 0.7 for the efficiency. Follow-through is allowed from part b).
Marker: calculates the energy efficiency of the platform lift
Each phrase of the question gave you something
QCAA did not publish a band for this question. It reads as simple familiar because QCAA split it into three short parts, each one formula long, and handed you each new number just before the part that needs it.
If I had to guess where you might lose marks, it would be part a). The stem front-loads the worm, the single thread and the 60 teeth, and a lot of students will reach for $\text{VR}=60$ when all part a) needs is the circumference. In part c), the trap is dividing 280 by 196 and getting an efficiency above 100%.
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 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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