Topic guide 4 · Unit 4 Topic 2 · Machines, mechanisms and control
Work, power and energy are where the rest of Unit 4 turns into an answer. Many long calculations on the paper are machine problems that finish with a work, a power or an efficiency, so the relationships in this guide turn up all through Section 3.
Here you will learn to find work and power, to move between kinetic, potential and total mechanical energy, and to run an efficiency in the right direction. Tick off each skill once you can do it without help.
Part 1
This topic ties Section 3 together. From 2020 to 2025, questions led by energy carried 36 Section 3 marks, and a work, power or efficiency step appeared in questions worth about 90 Section 3 marks. There was at least one multiple choice question on it every year, nine in all.
2023 leaned on it hardest. Questions 16, 17 and 21 were worth 24 marks between them.
Most of these questions are not purely about energy. They are machine or motion problems that finish with a power or an efficiency. For the machine half, use Topic guide 2 and Topic guide 3. This guide covers the energy half.
Unit 4 Topic 2, Machines, mechanisms and control, has three dot points for this topic:
The words “distance moved in direction of force” matter. A force at right angles to the motion does no work, which is why the normal force never appears in a work calculation.
Part 2
Work is the force multiplied by the distance moved in the direction of that force, $W = Fs$, measured in joules. If nothing moves, no work is done, however hard you push.
To use $W = Fs$ on a lifted load, the force is the weight of the load, $F = mg$. Weight is not on the formula sheet, so write it down yourself. Lifting a mass $m$ through a height $h$ at constant velocity takes work $W = mgh$, which is the potential energy the load gains.
Power is the rate of doing work, $P = W \div t$, measured in watts. One watt is one joule per second. A more powerful machine does the same job in less time. It does not do more work.
For something moving at a steady speed, $P = Fv$. It comes from $W = Fs$ and $P = W \div t$, but the combined form is not printed, so show where it comes from.
Work and energy are the same quantity, and both are measured in joules. The work done lifting something equals the potential energy it gains. The work done stopping something equals the kinetic energy it loses. Questions move freely between the two.
Practise
Each wrong option in these questions comes from a specific mistake, and the solutions name it.
The work done by a jack to vertically raise a 1500 kg vehicle a distance of 180 cm is
The force is the weight of the vehicle, and 180 cm is 1.8 m:
$W = mgh = 1500 \times 9.8 \times 1.8$
$W = 26\,460\ \text{J}$
(A) leaves out $g$, so it treats the mass as a force. (B) divides by the height instead of multiplying. (D) divides by 0.18 m, so the height is both converted wrongly and divided.
A hydraulic jack connected to a motor is used to raise an entire car. The mass of the car is 1500 kg. The motor uses 1.2 kW of power to raise the car.
What is the time taken to raise the car 500 mm vertically, rounded to two decimal places?
Find the work first. 500 mm is 0.5 m and 1.2 kW is 1200 W:
$W = mgh = 1500 \times 9.8 \times 0.5 = 7350\ \text{J}$
$t = \dfrac{W}{P} = \dfrac{7350}{1200} = 6.125 \approx 6.13\ \text{s}$
(A) is $1200 \div (1500 \times 0.5)$, which turns the formula upside down and drops $g$. (C) uses $g = 10$, which QCAA never uses. (D) leaves out the height.
Part 3
Potential energy is energy stored by height, $\text{PE} = mgh$. Kinetic energy is energy of movement, $\text{KE} = \tfrac{1}{2}mv^2$. Total mechanical energy is the sum of the two:
$$\text{ME} = \text{KE} + \text{PE}$$
That last formula is not in the formula book. The syllabus names it in words, “total mechanical energy is the sum of kinetic energy and potential energy”, and that is all the notice you get. 2023 Question 16 and 2024 Question 21 both turn on it.
Energy converts from one form to another without disappearing. For an object falling or launched with no friction, the kinetic energy it gains equals the potential energy it loses:
$$mgh = \tfrac{1}{2}mv^2 \Rightarrow v = \sqrt{2gh}$$
The mass cancels, so you do not need it. This gets you a speed from a height without the motion equations. The mark guides accept the motion route too, and it gives the same answer, but the energy route is shorter and needs no acceleration.
The QCAA likes to give you an energy so that you have to work backwards to a mass or a speed. In 2023 Question 16 a kinetic energy gives you the mass of a container. In 2024 Question 21 a total mechanical energy gives you the speed of a lift. Many past questions read a formula in reverse, including 2020 Q19a, 2023 Q16, 2023 Q21 and 2024 Q21.
Try it yourself
Set the height, the mass and the efficiency. Watch the potential energy turn into kinetic energy on the way down, and see how much friction takes. This model takes friction in step with the height lost, so the heat bar grows steadily on the way down.
Practise
For the written questions, each step shows the mark it earns, based on the QCAA marking guide. Follow-through is allowed, so if an early value is wrong but you use it correctly, you still earn the later marks.
A fibreglass swimming pool has a potential energy of 98 kJ when raised 20 m from the ground by a crane.
The mass of the swimming pool is
Read $\text{PE} = mgh$ in reverse. 98 kJ is 98 000 J:
$m = \dfrac{\text{PE} }{gh} = \dfrac{98\,000}{9.8 \times 20} = 500\ \text{kg}$
(B) leaves out $g$. (C) leaves out the height. (D) multiplies by $g$ instead of dividing.
An object has a mass of 1500 kg. It starts from rest and accelerates uniformly at 0.4 m/s² for a horizontal distance of 20 m.
What is the object’s kinetic energy?
Find $v^2$ from the motion equations, with $u = 0$:
$v^2 = u^2 + 2as = 0 + 2 \times 0.4 \times 20 = 16$
$\text{KE} = \tfrac{1}{2}mv^2 = \tfrac{1}{2} \times 1500 \times 16$
$\text{KE} = 12\,000\ \text{J} = 12\ \text{kJ}$
(B) leaves out the ½. (C) treats $v^2 = 16$ as $v = 16$ and squares it again. (D) is $mgh$ using the 20 m, but the object moves horizontally, so it gains no potential energy.
A crane requires 4355 W to vertically lift a full scrap metal bin a distance of 20 m from the ground at constant velocity over a period of 90 s.
a) Determine the mass of the full scrap metal bin to the nearest whole unit. [3 marks]
b) Determine the velocity on impact of a 5 kg section of scrap metal that falls to the ground unobstructed from the bin at the top of the lift. Answer to the nearest whole unit. [3 marks]
(a) Find the work from the power and the time:
$P = \dfrac{W}{t} \Rightarrow W = 4355 \times 90$
$W = 391\,950\ \text{J}$
1 mark · correct working for the work done.
The bin rises at constant velocity, so the forces balance and the work done equals the potential energy gained:
$391\,950 = m \times 9.8 \times 20$
1 mark · correct formula and substituted values.
$m = \dfrac{391\,950}{9.8 \times 20} = 1999.74 \approx 2000\ \text{kg}$
The mark guide prints 1999.75 here. The division gives 1999.74, and both round to 2000 kg.
1 mark · the answer to the nearest whole unit, with the unit.
(b) The piece falls freely, so its kinetic energy on impact equals its potential energy at the top:
$\text{KE} = mgh = 5 \times 9.8 \times 20 = 980\ \text{J}$
1 mark · correct working for the kinetic energy.
$\text{KE} = \tfrac{1}{2}mv^2 \Rightarrow v^2 = \dfrac{980}{0.5 \times 5} = 392$
1 mark · correct formula and substituted values.
$v = \sqrt{392} = 19.80 \approx 20\ \text{m s}^{-1}$
1 mark · the answer to the nearest whole unit, with the unit.
The mark guide also accepts the motion equations for (b), which give the same answer. Notice that (b) does not use your answer to (a). The falling piece is the 5 kg given in the question, and the mass of the bin does not matter. Past questions often look like one long chain but are two short ones, so read what each part needs before you assume the first answer feeds the second.
A crane lowers a storage container off the back of a truck at a velocity of 1.5 m/s, giving the container a kinetic energy of 850 J.
Calculate the total mechanical energy of the lowering system when the base of the storage container is 2 m above the ground. Answer to the nearest whole unit.
The kinetic energy gives you the mass:
$m = \dfrac{2 \times \text{KE} }{v^2} = \dfrac{2 \times 850}{1.5^2} = 755.56\ \text{kg}$
2 marks · the formula with values substituted, then the mass.
Potential energy with the base 2 m above the ground:
$\text{PE} = mgh = 755.56 \times 9.8 \times 2$
$\text{PE} = 14\,808.98\ \text{J}$
2 marks · the formula with values substituted, then the potential energy.
Add the two:
$\text{ME} = \text{KE} + \text{PE} = 850 + 14\,808.98$
$\text{ME} = 15\,658.98 \approx 15\,659\ \text{J}$
1 mark · the answer in joules to the nearest whole unit.
The final mark depends on ME = KE + PE, which is not in the formula book. The question also gives you an energy so that you have to work backwards to a mass. The mark guide accepts a potential energy from 14 808.78 J to 14 818 J, so keep at least two decimal places in the mass.
A lift has a total mass of 1000 kg. The lift undergoes uniform acceleration from a stationary position on the ground floor until it reaches a height of 10 m and has a total mechanical energy of 106 kJ. From this point it continues to travel vertically at a constant velocity.
Determine the time taken for the lift to reach its constant velocity.
This question joins energy to motion. The energy half gets you a speed, and the motion half turns that speed into a time. Neither half works on its own.
Potential energy at 10 m.
$\text{PE} = mgh = 1000 \times 9.8 \times 10$
$\text{PE} = 98\,000\ \text{J} = 98\ \text{kJ}$
2 marks · seeing that the potential energy comes first, then calculating it.
The kinetic energy is what is left.
$\text{ME} = \text{KE} + \text{PE}$
$\text{KE} = 106 - 98 = 8\ \text{kJ} = 8000\ \text{J}$
2 marks · recognising that total mechanical energy is the sum of the potential and kinetic energy, then calculating the kinetic energy.
Speed from the kinetic energy.
$8000 = \tfrac{1}{2} \times 1000 \times v^2$
$v^2 = 16 \Rightarrow v = 4\ \text{m s}^{-1}$
2 marks · seeing that the speed comes from the mass and the kinetic energy, then calculating it.
Acceleration. The lift starts from rest and covers 10 m:
$v^2 = u^2 + 2as \Rightarrow 16 = 0 + 2a(10)$
$a = 0.8\ \text{m s}^{-2}$
2 marks · the right motion formula, then the acceleration.
Time.
$v = u + at \Rightarrow 4 = 0 + 0.8t$
$t = 5\ \text{s}$
2 marks · the right motion formula, then the time.
The mark guide also accepts $s = ut + \tfrac{1}{2}at^2$ for the time, and it allows follow-through throughout. An error in the kinetic energy can still earn all four motion marks.
Part 4
Efficiency is the useful output divided by the input:
$$\eta = \frac{\text{useful output} }{\text{input} } \times 100\%$$
It works with energies or with powers, as long as the top and the bottom are the same kind. The formula book prints three versions, $\eta = \dfrac{\text{MA} }{\text{VR} } \times 100\%$, $\eta = \dfrac{P_{out} }{P_{in} } \times 100\%$ and $\eta = \dfrac{E_{out} }{E_{in} } \times 100\%$. Use the one that matches what you were given. The rearrangement you will need most often is:
$$\text{input} = \frac{\text{output} }{\eta}$$
The output is always smaller than the input. Going forwards, with the input known and the output wanted, multiply by the efficiency. Going backwards, with the output known and the input wanted, divide by it.
If your input comes out smaller than your output, you multiplied when you should have divided. Backwards is always input = output ÷ η. A machine cannot give out more than it takes in.
For example, a machine gives 300 J of useful output at 75% efficiency. The input is $300 \div 0.75 = 400$ J. Multiplying instead gives $300 \times 0.75 = 225$ J, which is smaller than the output, so it cannot be right.
The 2023 subject report names one efficiency error: treating 80% efficiency as 100% plus a 20% loss and multiplying by 1.2. That answer is bigger than the output, so the check above will not catch it.
In 2023 Question 21 the first pump outputs 490 W. The correct input is $490 \div 0.8 = 612.5$ W. Multiplying by 1.2 gives 588 W, which is wrong.
When one machine drives another, so the output of the first is the input of the second, the overall efficiency is the product of the two. It is never the average. Efficiencies of 90% and 80% give $0.9 \times 0.8 = 0.72$, or 72% overall. This rule is not on the formula sheet.
Not every two-stage question works this way. In 2023 Question 21 each pump takes its own power and lifts the water through its own height. There you work out each pump separately, divide each output by its own efficiency, and add the two inputs. Multiplying the efficiencies gives the wrong answer.
Practise
Each step shows the mark it earns, from the 2023 marking guide.
A factory uses a two-pump system to vertically lift water into a main reservoir.
The first pump is 80% efficient and lifts 500 litres of water per minute into a holding tank 6 m above the ground. The second pump is 75% efficient and lifts 500 litres of water per minute from the holding tank into the main reservoir 10 m above the ground.
This current system is being replaced with a single-pump system that is 90% efficient and has the same power input as the two-pump system. Determine the rate at which the new system will lift water into the main reservoir. Answer to the nearest whole unit (litres per minute).
Note: 1 litre of water has a mass of 1 kg.
Start with the height each pump lifts through. The holding tank is 6 m up and the reservoir is 10 m up, so the second pump raises the water only 4 m. The replacement pump lifts from the ground to the reservoir, so it uses the full 10 m.
First pump. 500 L is 500 kg, lifted 6 m each minute:
$W = mgh = 500 \times 9.8 \times 6 = 29\,400\ \text{J}$
1 mark · the work done by the first pump.
$P_{out} = \dfrac{29\,400}{60} = 490\ \text{W}$
1 mark · the power output of the first pump.
$P_{in} = \dfrac{490}{0.8} = 612.5\ \text{W}$
1 mark · the power input of the first pump.
Second pump, through 4 m:
$W = 500 \times 9.8 \times 4 = 19\,600\ \text{J}$
1 mark · the work done by the second pump.
$P_{out} = \dfrac{19\,600}{60} = 326.67\ \text{W}$
1 mark · the power output of the second pump.
$P_{in} = \dfrac{326.67}{0.75} = 435.56\ \text{W}$
1 mark · the power input of the second pump.
The current system.
$P_{in} = 612.5 + 435.56 = 1048.06\ \text{W}$
1 mark · the overall power input of the two-pump system.
The new pump has the same input and is 90% efficient. Going forwards, multiply:
$P_{out} = 0.9 \times 1048.06 = 943.25\ \text{W}$
1 mark · the power output of the new pump.
$W = P \times t = 943.25 \times 60 = 56\,595\ \text{J}$
1 mark · the work done by the new pump each minute.
$m = \dfrac{W}{gh} = \dfrac{56\,595}{9.8 \times 10} = 577.5\ \text{kg}$
1 mark · the answer in litres per minute to the nearest whole unit.
577.5 kg is 577.5 L, so the new pump lifts 578 litres per minute. The mark guide accepts 577 to 578.
Using 10 m for the second pump is the trap in this question. It inflates the answer by about 60%, to 938 litres per minute instead of 578.
The 2023 mark guide allows follow-through at every step, including “for incorrect power input of either or both pumps when determining the overall power input”. A mistake early in the chain costs you one mark, provided you carry the wrong number correctly through every later step.
Part 5
The table lists the past QCAA questions that use this topic. Items whose main topic is something else, but which finish with a work, power or efficiency step, show that main topic in italics. Band tags are derived, except for 2020, where the QCAA published the split, and for the items the 2021 and 2022 subject reports name.
Past QCAA questions: work, power, energy and efficiency
| Question | Marks · band | What it tests |
|---|---|---|
| 2020 Q6 (MC) | 1 · SF | Efficiency of an irrigation pump. Answer 37%. |
| 2021 Q9 (MC) | 1 · SF | Work done by a jack raising 1500 kg through 180 cm. Answer 26 460 J. Worked above. |
| 2022 Q3 (MC) | 1 · SF | Power a go-kart’s wheels transmit to the track, $P = Fv$. Answer 1.5 kW. |
| 2022 Q5 (MC) | 1 · SF | Potential energy of a full water tank, measured at its midpoint. Answer 1.47 MJ. |
| 2023 Q10 (MC) | 1 · SF | Work done pushing 40 kg 3 m up an incline at uniform velocity. Answer 986 J. |
| 2024 Q8 (MC) | 1 · SF | Mass of a swimming pool from 98 kJ of potential energy at 20 m. Answer 500 kg. Worked above. |
| 2024 Q10 (MC) | 1 · SF | Work done raising a box through a pulley system. Answer 360 J. |
| 2025 Q8 (MC) | 1 · SF | Kinetic energy after uniform acceleration from rest. Answer 12 kJ. Worked above. |
| 2025 Q10 (MC) | 1 · SF | Time for a 1.2 kW motor to raise a 1500 kg car 500 mm. Answer 6.13 s. Worked above. |
| 2020 Q19 | 6 · CF | Crane. Mass from power and time, then impact velocity from a fall. Worked above. |
| 2020 Q20 | 9 · CU | Worm winch, then compound-machine efficiency. Gears. |
| 2021 Q16 | 5 · SF | Windlass at a well. Explain the work done and the power used, with mathematical reasoning (Section 2). |
| 2021 Q20 | 5 · SF | Piledriver. Velocity just after impact once 5.4 kJ of the 19.6 kJ is lost (3 marks), then the stopping distance (2 marks). |
| 2021 Q22 | 6 · CF (est.) | Winch crane output power, then motor revolutions. Gears. |
| 2022 Q20 | 6 · SF | Generator pulled up a 15° incline by an 80% efficient pulley system. Friction and inclines. |
| 2023 Q16 | 5 · SF (est.) | Storage container. Mass from KE, then total mechanical energy. Worked above. |
| 2023 Q17 | 9 · SF (est.) | Threaded rod. Power required from a 62% efficient motor. Screws. |
| 2023 Q18 | 7 · SF (est.) | Electric bicycle, finishing with motor output power. Gears. |
| 2023 Q21 | 10 · CU (est.) | Two-pump water system replaced by one pump. Answer 578 L/min. Worked above. |
| 2024 Q18 | 5 · SF (est.) | Screw jack. Work on the lever arm, then power input. Screws. |
| 2024 Q21 | 10 · CU (est.) | Lift. Total mechanical energy to velocity to time. Worked above. |
| 2025 Q17 | 3 · SF (est.) | Block and tackle at 100% efficiency. Pulleys. |
| 2025 Q18 | 4 · SF (est.) | Bottle opener: load force, then work done. Levers. |
| 2025 Q20 | 5 · SF (est.) | Worm-driven platform lift. Turns, work, then efficiency. Gears. |
The 2020 to 2025 papers have four ten-mark questions. Three are calculations: 2021 Q23 (friction and motion), 2023 Q21 and 2024 Q21. The fourth, 2022 Q16, is logic. So two of the three ten-mark calculations are energy chains.
2023 Q21 closes its paper. 2024 Q21 is followed by an eight-mark incline question. Here is how each paper finished:
| Year | Final question | Marks |
|---|---|---|
| 2020 | Q23 lift (Newton’s second law) | 6 |
| 2021 | Q23 conveyor, static friction | 10 |
| 2022 | Q22 luggage system, friction then rpm | 9 |
| 2023 | Q21 pumps (energy chain) | 10 |
| 2024 | Q22 incline | 8 |
| 2025 | Q22 sled on an incline | 8 |
In five of the six papers the last question was worth 8 to 10 marks. Four of those five were friction, incline or motion chains, and one was an energy chain. Leave enough time to work the last question in steps instead of hunting for a shortcut. Neither energy chain is conceptually hard, but both are long.
Part 6
| Checkpoint | How to run it |
|---|---|
| Is the input larger than the output? | It always is. If not, you applied the efficiency the wrong way round. |
| Did I divide by η? | 80% means ÷ 0.8, never × 1.2. The 2023 subject report flags this error. |
| Which height does this stage lift through? | A second-stage pump or lift raises the load only the remaining distance. |
| Work or power? | Joules or watts. If the question gives a time, it probably wants a power. |
| Did the mass cancel? | In $mgh = \tfrac{1}{2}mv^2$ it does. If you are still carrying it, check your working. |
| Is the force in the direction of motion? | A force at right angles to the motion does no work. |
| Have I converted? | Centimetres and millimetres to metres, kilojoules to joules, litres of water to kilograms, minutes to seconds. |
| Did I answer with a unit? | Joules, watts, metres per second or kilograms. |
Keep at least two decimal places in every intermediate value and round only the final answer. Give it to the nearest whole unit unless the question states a precision. The 2023 subject report recommends leaving rounding until the end, because early rounding can push an answer outside the accepted range.
The unit is usually part of the final-answer mark, and occasionally it is a mark of its own. Leave it off and you lose that mark.
The mechanical dot points are substantively the same. The 2019 syllabus wrote work as $W = Fd$. The 2025 syllabus writes $W = Fs$, with $s$ the “distance moved in direction of force”. Version 1.4 of the syllabus (March 2026) added “× 100%” to the efficiency formulas, which settles whether the formula gives a decimal or a percentage.
Two electrical efficiency formulas now sit alongside the mechanical one, $\eta = \dfrac{P_{out} }{P_{in} } \times 100\%$ and $\eta = \dfrac{E_{out} }{E_{in} } \times 100\%$. They are the same relationship applied to circuits, so efficiency now runs through both halves of Topic 2. Topic guide 8 covers them.
Game plan
Game plan 2 hands over to this topic at $W = Fs$, and Game plan 1 hands over at $\tfrac{1}{2}mv^2$.
Practice
Four questions written to QCAA form, with every number checked. The mark notes follow the style of the QCAA marking guides. Final answers are to the nearest whole unit unless the question says otherwise.
A motor raises an 80 kg load 12 m vertically in 15 seconds. What is the output power of the motor?
$W = mgh = 80 \times 9.8 \times 12 = 9408\ \text{J}$
$P = \dfrac{W}{t} = \dfrac{9408}{15} = 627.2\ \text{W}$
(A) leaves out $g$, so it treats the mass as a force. (C) divides the work by the height instead of the time. (D) is the work in joules with a watt label, so the time was never used. A unit check catches (D).
A hoist raises a 250 kg load 4 m vertically in 8 seconds. The hoist is 65% efficient.
a) Calculate the work done on the load. [1 mark]
b) Calculate the useful power output of the hoist. [1 mark]
c) Calculate the power input required. [2 marks]
(a)
$W = mgh = 250 \times 9.8 \times 4 = 9800\ \text{J}$
1 mark · correct formula and substituted values to give the work done.
(b)
$P_{out} = \dfrac{W}{t} = \dfrac{9800}{8} = 1225\ \text{W}$
1 mark · the useful power output.
(c) The output is known and the input is wanted, so divide:
$P_{in} = \dfrac{1225}{0.65} = 1884.62 \approx 1885\ \text{W}$
2 marks · recognising that the input is the output divided by the efficiency, then the answer with its unit.
Follow-through applies, so a wrong work value in (a) can still earn the marks in (b) and (c).
A 12 kg parcel leaves a chute travelling at $6\ \text{m s}^{-1}$ when its base is 3.5 m above the floor.
a) Calculate the total mechanical energy of the parcel at that point. [3 marks]
b) Determine the speed of the parcel as it reaches the floor, assuming no energy is lost. [2 marks]
(a)
$\text{KE} = \tfrac{1}{2}mv^2 = \tfrac{1}{2}(12)(6)^2 = 216\ \text{J}$
1 mark · correct formula and substituted values for the kinetic energy.
$\text{PE} = mgh = 12 \times 9.8 \times 3.5 = 411.6\ \text{J}$
1 mark · correct formula and substituted values for the potential energy.
$\text{ME} = 216 + 411.6 = 627.6 \approx 628\ \text{J}$
1 mark · recognising that ME = KE + PE, then the total.
(b) At floor level all of the mechanical energy is kinetic. Carry the unrounded 627.6 J:
$627.6 = \tfrac{1}{2}(12)v^2 \Rightarrow v^2 = 104.6$
1 mark · recognising that all the energy is kinetic at the floor.
$v = 10.23 \approx 10\ \text{m s}^{-1}$
1 mark · the speed, with its unit.
The motion route gives the same speed, $v = \sqrt{u^2 + 2gh} = \sqrt{36 + 2(9.8)(3.5)} = 10.23\ \text{m s}^{-1}$, and earns full marks.
A grain-handling plant uses two conveyors in series. The first is 70% efficient and raises 600 kg of grain per minute to a hopper 5 m above the ground. The second is 60% efficient and raises the same 600 kg per minute from the hopper to a silo inlet 8 m above the ground.
The plant is replacing both with a single 85% efficient conveyor that lifts grain from the ground directly to the silo inlet, using the same total power input.
Determine the mass of grain the new conveyor will raise each minute.
This question follows 2023 Question 21. Allow about twelve minutes.
The second conveyor raises the grain only the remaining 3 m, but the new conveyor lifts it the whole 8 m. Work in power once you reach the second stage, and keep track of which figures are per minute.
Conveyor 1, through 5 m:
$W = 600 \times 9.8 \times 5 = 29\,400\ \text{J per minute}$
$P_{out} = \dfrac{29\,400}{60} = 490\ \text{W}$
1 mark · the work done and power output of the first conveyor.
$P_{in} = \dfrac{490}{0.70} = 700\ \text{W}$
1 mark · the power input of the first conveyor.
Conveyor 2, through the remaining 3 m:
$W = 600 \times 9.8 \times 3 = 17\,640\ \text{J per minute}$
1 mark · recognising that the second conveyor raises the grain through 3 m.
$P_{out} = \dfrac{17\,640}{60} = 294\ \text{W}$
$P_{in} = \dfrac{294}{0.60} = 490\ \text{W}$
1 mark · the power output and input of the second conveyor.
Combined input:
$700 + 490 = 1190\ \text{W}$
1 mark · the overall power input of the current system.
New conveyor, same input, 85% efficient, full 8 m lift:
$P_{out} = 0.85 \times 1190 = 1011.5\ \text{W}$
1 mark · the power output of the new conveyor.
$W = 1011.5 \times 60 = 60\,690\ \text{J per minute}$
1 mark · the work done by the new conveyor each minute.
$m = \dfrac{W}{gh} = \dfrac{60\,690}{9.8 \times 8} = 774.11 \approx 774\ \text{kg}$
1 mark · the mass per minute, with its unit.
Using 8 m for the second conveyor gives a combined input of 2006.67 W and 1305 kg per minute. The arithmetic is consistent, but the second conveyor only lifts the grain 3 m.
Sources: QCAA Engineering external assessment question and response books, multiple choice question books and marking guides 2020 to 2025 · Engineering 2025 v1.4 general senior syllabus, Unit 4 Topic 2 · QCAA Engineering subject reports 2020 to 2025 · Engineering formula and data book 2025 v1.4. All numerical answers verified computationally against the published marking guides.
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