Question 23
An elevator starts from rest and accelerates upwards at a constant rate until the desired velocity of 10 m/s is achieved. The mass of the elevator and occupants is 2500 kg. The tension on the elevator cable is 30 000 N.
Determine the distance the elevator travels during acceleration to the nearest whole unit (m). Include a force diagram with your working.
Only the tension left over after the weight accelerates the elevator
The force arrows are drawn to scale. Compare the tension with the weight, then watch the elevator rise while its velocity-time graph draws itself, slowed down a little.
The cable has to hold up all 24 500 N of weight before it can accelerate anything. Only the 5500 N left over goes into $F=ma$.
Change the tension, the load and the target speed
Set the cable tension, the mass of the elevator and occupants, and the velocity it accelerates to. The question is 30 000 N, 2500 kg and 10 m/s. Switch the weight off to see the most expensive mistake.
This is the question. The tension beats the weight by 5500 N, which gives $a=2.20\text{ m/s}^2$. Reaching 10 m/s takes 4.55 s, and the elevator rises 22.7 m, which is 23 m to the nearest whole unit.
With the weight left out, all of the tension seems to accelerate the elevator. The acceleration comes out far too big and the distance far too small. With the question’s numbers you would write 4 m instead of 23 m.
The tension is no bigger than the weight, so the net force is zero or points down. An elevator starting from rest cannot accelerate upwards, and the question has no answer.
A small change in tension makes a big change in the net force, because the weight takes most of it. Halve the net force and the distance doubles. Both routes always agree, apart from the rounding of $t$.
Work through the solution one mark at a time
Try each step yourself before you reveal it.
Draw the elevator as a box with three labelled arrows: the tension up, the weight down, and the acceleration up beside it. The tension arrow is a little longer than the weight arrow, because the elevator accelerates upwards.
The acceleration arrow is part of the mark. It is not a force, so draw it off to the side of the box.
Marker: provides an appropriate force diagram. Note: “An appropriate force diagram includes correctly applied force vectors for tension, weight and acceleration.”
Find the net force, the force that is left to accelerate the elevator. It is the tension minus the weight.
$$\begin{aligned}ma&=T-mg\\&=30\,000\\&\quad-(2500\times9.8)\\&=5500\text{ N}\end{aligned}$$
The guide writes the weight as $W$. In the 2026 formula book $W$ is work, so write $mg$ to keep the two apart.
Marker: provides correct working to give correct force required for acceleration
Divide the net force by the mass.
$$a=\frac{5500}{2500}=2.20\text{ m/s}^2$$
Marker: provides correct working to give correct acceleration
The sample response finds the time to reach 10 m/s next, starting from rest.
$$\begin{aligned}v&=u+at\\10&=0+(2.20\times t)\\t&=\frac{10}{2.20}=4.55\text{ s}\end{aligned}$$
Marker: provides correct working to give correct time to complete acceleration. Note: “Correct time to complete acceleration range: 4.545 454 545 to 4.55 s”
Write the motion equation for the distance, then write it again with every number in.
$$\begin{aligned}s&=ut+\tfrac12at^2\\&=(0\times4.55)\\&\quad+\tfrac12(2.2\times4.55^2)\end{aligned}$$
Marker: provides correct formula and substituted values. Note: “Formula is considered correct if substituted and constant values are correctly applied.”
Evaluate, then round to the nearest whole metre and give the unit.
$$s=22.77\text{ m}$$
The 22.77 comes from the rounded time of 4.55 s. Kept unrounded, the time gives 22.73 m. The guide also accepts going straight to $v^2=u^2+2as$, which skips the time.
$$\begin{aligned}s&=\frac{v^2-u^2}{2a}=\frac{10^2-0^2}{2\times2.2}\\&=22.73\text{ m}\approx23\text{ m}\end{aligned}$$
The guide prints that alternative as “$v^2+u^2+2as$”. It means $v^2=u^2+2as$. Follow-through is allowed at every stage, so a wrong net force, acceleration or time still earns the later marks if you use it correctly.
Marker: provides correct working to give correct answer to the nearest whole unit (m) with correct unit provided. Note: “Allow FT error for incorrect: acceleration force in determining acceleration, acceleration in determining time to complete acceleration, time to complete acceleration in determining the answer.”
Each phrase of the question gave you something
Each step is familiar, but you have to chain three of them yourself: a net force you must build from two forces, an acceleration, and then a motion equation. The 2020 subject report lists two complex familiar questions worth 12 marks in Section 3, and Questions 19 and 23 are the only 6-mark questions in that section, so the band is QCAA’s own.
If I had to guess where you might lose marks, it would be using the 30 000 N as the net force. That gives $a=12\text{ m/s}^2$ and a distance of 4 m. It costs you the force mark at once, and every later mark then depends on follow-through. The other easy loss is the force diagram without the acceleration arrow, which the guide names as part of an appropriate diagram.
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 2020 QCAA Engineering external assessment, © State of Queensland (QCAA) 2020, 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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