Question 18
A bottle opener is used to remove the bottle cap from a glass drink bottle. An effort force of 20 N is exerted on the bottle opener to lift the bottle cap by 2.5 mm. The bottle opener has a velocity ratio (VR) of 16 and is 90% efficient.
Your hand moves 16 times as far as the cap, but friction keeps the force gain to 14.4
The opener is a lever with the cap between the fulcrum and your hand. The arms are drawn in the ratio 16 to 1 and the two lifts are to scale. The force bars at the bottom share one scale.
The VR comes from the geometry, so friction cannot touch it. The efficiency shrinks only the force you get out, from 320 N to 288 N.
Change the opener, then try the slips that cost marks
The dashed lever is the starting position and the solid one is lifted. The question is 20 N, VR 16, 90% and a 2.5 mm lift.
This is the question. $\text{MA}=0.9\times16=14.4$, so the load force is 288 N. Your hand moves $16\times2.5=40$ mm, so the work done on the opener is $20\times0.04=0.8$ J. The load side gives the same 0.8 J once you divide its 0.72 J by 0.9.
The VR sets how far your hand moves. The efficiency only shrinks the force you get out. Push the efficiency to 100% and the load force becomes VR times the effort, which is the only time you may write $\text{MA}=\text{VR}$.
Using $\text{MA}=\text{VR}$ pretends the opener is perfect. That is only true at 100%, and this one is 90%, so the load force comes out too big.
The load force with the effort distance belongs to neither side of the machine, so the work is too big by a factor of the MA, 14.4 at the question’s values. Pair the effort force with the effort distance, or the load force with the load distance.
The guide accepts the effort distance in millimetres as an answer to that step, but a joule is a newton metre. Put millimetres into $W=Fs$ and the work comes out a thousand times too big.
Work through the solution one mark at a time
Try each step yourself before you reveal it.
Start from the formula book form of efficiency and rearrange it for MA, with $\eta$ as a decimal.
Write the rearranged line, $\text{MA}=\eta\times\text{VR}$, before the numbers. The 2025 subject report asks for exactly this kind of explicit statement.
Marker: correctly calculates MA
MA is the force ratio, so the load force is MA times the effort.
Follow-through is allowed, so a wrong MA from step 1 still earns this mark if you multiply by it correctly.
Marker: calculates load force on the bottle opener
The work done on the opener is done by your hand, so you need the effort distance. The stem gives the load distance, so multiply by the VR.
Marker: calculates distance of effort. “Accept distance of effort in mm or m.”
Pair the effort force with the effort distance, in metres.
$$W=F_E\,d_E=20\times0.04$$
The guide also accepts the load side: $W_L=F_L\,d_L=288\times0.0025=0.72$ J is the useful work on the cap, and $W_E=\tfrac{0.72}{0.9}=0.8$ J. The guide writes $W=Fd$, which is the formula book’s $W=Fs$.
Leave the answer as 0.8 J. The guide’s front page says final answers go to the nearest whole unit unless stated, which would make it 1 J, but its own sample answer is 0.8 J and the 2025 subject report says not to round to a whole unit when the question does not ask you to.
Marker: calculates work done on the bottle opener to lift the bottle cap
Each phrase of the question gave you something
The 2025 subject report says students did well on the simple familiar calculations that found the mechanical advantage and the load force from a given efficiency, VR and effort force, which is part a) exactly. It also asks for explicit statements of the working in every calculation, and gives MA = VR when 100% efficient as its example.
QCAA did not publish a band for this question by number. It reads as simple familiar because every value is given in the stem and each part is two formula book steps, and the 2025 subject report names the part a) calculation in its list of simple familiar work.
If I had to guess where you might lose marks, it would be part b). The stem hands you 2.5 mm, and $20\times0.0025=0.05$ J looks like an answer. It pairs the effort force with the load distance, so it is wrong by a factor of 16.
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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