Fast answer: Work is an amount of energy transferred, power is the transfer rate, and efficiency is the useful-output share of the total energy input. This guide explains how to tell them apart in motorised-ramp experiments, calculate each correctly and resist claims that the observations do not support.
Start here: Read the definitions and full lift calculation, then choose a practice case that exposes your first error. The official 2027 SEC G3 Physics syllabus is K323; the examples below are original practice, not examination questions.
Contents
- A motor is fast; does that mean it has used more energy?
- The quick answer you can use immediately
- The actual SEC examination, not a guessed paper
- Draw the energy boundary before doing arithmetic
- A complete motor lift calculation
- Same energy, a different time
- Power from a force is not always force times speed
- Efficiency needs a stated definition of useful
- Measurements and the limits of the apparatus
- Graph language: separate slope, trend and explanation
- What a results table can establish
- Recover when the question is worded indirectly
- Common mistakes that cost easy marks
- A seven-day practice route
- Practice case 1 — calculate only what is supported
- Practice case 2 — separate operating speed from efficiency
- Practice case 3 — the interval on a graph matters
- Practice case 4 — account for speed as well as height
- Practice case 5 — repair an inverted ratio
- Practice case 6 — compare rated and measured quantities
- Practice case 7 — make a fair test
- Practice case 8 — read a conclusion before accepting it
- Practice case 9 — choose the right denominator
- Practice case 10 — inverse calculation from efficiency
- Practice case 11 — write a data-based explanation
- Practice case 12 — an unfamiliar context
- How to answer a structured examination question well
- Frequently asked questions
- What to read next
A motor is fast; does that mean it has used more energy?
A motorised trolley rises through a measured height on a laboratory track. A learner hears that the faster motor must have done more work. Another student argues that a motor which finishes the same climb more quickly must be more efficient. Neither conclusion follows from speed alone. The central distinction is between an amount of energy, the rate at which energy is transferred, and the fraction of supplied energy that reaches the intended output. These are three different questions, with three different calculations.
For a given mass and vertical height change, the increase in gravitational potential energy is mgh. A lift that raises the same trolley through the same height transfers the same amount into that gravitational store even when it takes a different time. If one lift takes less time, its average useful power is greater. To judge efficiency, however, we need an input measure as well as the useful output. Without an input figure, the word efficient is merely a guess.
This guide extends Volume 0683 on the trolley energy pathway and Volume 0684 on speed-squared graphs. Those guides establish where mechanical energy comes from and how a relationship may be represented. Our task here is to account for the rate of transfer and the denominator of efficiency, then to defend the conclusions in an examination answer.
The quick answer you can use immediately
Work done is energy transferred by a force acting through a displacement in the direction of that force. In the simple aligned case, work done = force × distance. Power is the rate of energy transfer: power = energy transferred ÷ time taken. Efficiency is useful energy output ÷ total energy input, commonly expressed as a percentage. All three expressions are quantities, not interchangeable labels. Work and energy are measured in joules, power in watts, and efficiency is a ratio without a physical unit.
Before substituting, name the system and the useful outcome. For a lifting motor, gravitational potential energy gained by a load may be the useful output. For a classroom ramp, a moving trolley may also gain kinetic energy, so the question may need a wider energy balance. The words ‘useful’ and ‘input’ depend on the task being studied: energy heating a motor housing is generally an unwanted transfer for a lift, but useful heating would be the objective of a heater.
Do these steps in order: establish the change, identify the input, decide whether the question asks for amount, rate or proportion, use consistent SI units, calculate without premature rounding, and return to the context. This prevents a correct arithmetic operation from answering the wrong question. It also works as an examination recovery route when the problem appears unfamiliar.
The actual SEC examination, not a guessed paper
The 2027 SEAB G3 Physics syllabus K323 is the official source for this guide. It replaces the 2026-and-earlier reference code 6091. Candidates take Paper 1 multiple choice, 1 hour, 40 marks, worth 30%; Paper 2 structured and free response, 1 hour 45 minutes, 80 marks, worth 50%; and Paper 3 practical, 1 hour 50 minutes, 40 marks, worth 20%. The physics energy section expressly covers stores, transfers, work, power and efficiency. Use the exact syllabus for your examination year rather than assuming a school worksheet is the examination specification.
The 2027 SEC is the first combined certificate framework, but G3 examination standards correspond to the prior O-Level standard. The new subject code changes the identity of the examination; it does not justify assuming a different physics law. Questions can require knowledge, numerical and graphical reasoning, and experimental evaluation. Being able to repeat a formula is only the starting point.
Good preparation therefore includes three kinds of rehearsal. First solve small quantities cleanly to secure units. Then face mixed information, deciding which quantities are actually needed. Finally explain what the answer permits a reader to conclude. A Paper 2 response may test all three in one structured question, while a practical discussion may use the same thinking without presenting the equation in its simplest form.
Draw the energy boundary before doing arithmetic
A useful energy account begins with a boundary. Imagine a motor, its electrical supply, a belt, and a trolley. Energy enters the motor electrically. Mechanical transfer reaches the trolley. The trolley may gain height and speed. Energy also reaches the surroundings by heating, sound and friction. The boundary determines which of those transfers are counted as useful and where the loss is recorded.
If the question only studies a load lifted slowly from rest to rest, gravitational potential energy gained is a straightforward measure of useful output. If the load accelerates, it also gains kinetic energy. Ignore that increase and the measured efficiency may be understated. If it is already moving at the start and finish with different speeds, read the values before deciding what belongs in useful output. There is no rule that the gravitational term is always the entire story.
Use a conservation sentence after your computation: the energy supplied is accounted for by changes in stored energy and transfers to the surroundings. Energy is not destroyed. The term ‘lost energy’ is an everyday shorthand for energy dispersed into less useful stores or surroundings. In examination work, name the pathway rather than writing that energy vanishes.
A complete motor lift calculation
Suppose a 2.0 kg trolley is raised vertically by 0.75 m and is stationary at both ends. Take gravitational field strength as 10 N/kg for this worked example. The useful output is mgh = 2.0 × 10 × 0.75 = 15 J. If the motor completes the lift in 3.0 s, its average useful power is 15 ÷ 3.0 = 5.0 W. If the measured electrical input during the lift is 24 J, the energy efficiency is 15 ÷ 24 × 100% = 62.5%.
Notice that three answers describe one event. Fifteen joules answers how much useful energy was gained. Five watts answers the average rate of useful transfer. A 62.5% efficiency answers what proportion of supplied energy became the chosen output. The motor may draw more than 5 W electrical input because the input must also account for transfers not classified as useful. For a steady average input of 24 J over 3.0 s, the average input power is 8.0 W.
The checking sentence is as valuable as the numbers: the useful output is below the recorded total input, and the ratio is between zero and one. The unexplained difference is 9 J, assigned within this simplified model to other energy transfers. We should not claim that all 9 J became sound unless the experiment has established where it went.
Same energy, a different time
Now compare two lifts with the same trolley and vertical rise. Motor A takes 3 s and Motor B takes 6 s. If both deliver 15 J to the load, average useful power is 5 W for A and 2.5 W for B. Motor A transfers the same useful energy in half the time. This tells us that its average useful power is twice as great. It does not say that Motor A used twice as much energy or that it wastes more.
Suppose Motor A consumes 24 J while B consumes 20 J. Their efficiencies become 62.5% and 75% respectively. The quicker motor in this example is less efficient. The facts are compatible because efficiency compares energy amounts, while power introduces time. A question can deliberately choose numbers that challenge the instinct to treat ‘fast’ and ‘efficient’ as the same word.
When writing a comparison, use paired sentences. A is more powerful on the useful-output basis because it completes the same energy transfer in less time. B is more energy-efficient because it requires a smaller total input for that same useful output. Avoid giving either sentence without the ‘because’ clause. The reason proves that the learner has selected the correct measure.
Power from a force is not always force times speed
The syllabus relationship work done = force × distance moved in the direction of the force applies to aligned motion, as presented at this level. For steady force and speed along the same direction, the average power over a time interval can be found from work divided by time, which leads to force × speed in that restricted situation. But the general question might include a changing force, a varying speed, an angled direction or an interval where energy enters more than one store.
Do not reach mechanically for force × speed whenever both appear. Ask whether the question actually gives a constant aligned force and the appropriate speed. A ramp provides a useful warning. The driving force acts up the slope while the gain in gravitational potential energy depends on vertical height, not the distance along the track. Work done against gravity and work done by a motor need not be the same if friction, acceleration or other effects matter.
If a student calculates 10 N × 0.8 m/s = 8 W while the question concerns a 10 N force at an angle to the motion, that value may be inappropriate. At examination level, use the data and diagram to establish the aligned component when relevant, or return to energy divided by time if that is what is supplied. The equation must describe the situation, not merely fit its numbers.
Efficiency needs a stated definition of useful
A motorised ramp can serve several purposes. If the purpose is to raise cargo, increased gravitational potential energy is useful. If the purpose is to launch a trolley at a specified speed, the desired kinetic energy may be useful. If the trolley is intended to arrive at a higher platform with a specified speed, both changes may matter. The same apparatus can have different useful outputs because the goal is different.
The total input must be on the same energy basis and for the same trial. Using the electrical energy from Tuesday with the height gained on Wednesday produces a precise-looking percentage that has no experimental meaning. Similarly, input power and output energy cannot be divided directly: watts divided into joules do not give an efficiency. Convert power into energy using the relevant duration, or compare powers over the same steady interval when justified.
A useful sentence to attach to the equation is: ‘I count the increase in the load’s gravitational potential energy as useful because the purpose is to lift the load.’ This makes the numerator explicit. Then identify the total energy supplied as the denominator. It takes only one extra line but protects the entire argument from hidden assumptions.
Measurements and the limits of the apparatus
An experiment needs more than a single reading if it is intended to compare motors. Record the load mass, vertical rise, duration, starting and ending motion state, input energy method, and relevant observations about friction or slipping. Repeat comparable trials and examine whether the readings are stable. If a motor overheats between trials, the order of testing may change its behaviour; the comparison should control or report that difference.
Be careful about measuring electrical input. A stated wattage on a motor label may be a rated value, not the actual energy consumed during one trial. Likewise, a stopwatch operated by hand introduces reaction-time uncertainty. A small time difference between motors is less persuasive when the timing uncertainty is large relative to that difference. The better answer is not to hide the issue but to identify a credible improvement, such as a sensor-based time measurement.
The measurement question can be written as a chain: what did we read, what quantity do we require, what conversion joins them, and which assumptions remain? For example, multiplying an average measured electrical input power by elapsed time provides an energy estimate when the average is valid. Do not present a single snapshot of instantaneous power as a measurement of the energy for the whole run.
Graph language: separate slope, trend and explanation
A plot of cumulative energy transferred against time can show a rate through its gradient. When the graph is a straight line through the origin, the constant gradient represents constant power. If the graph curves, power changes with time; the average power over an interval is the change in energy divided by elapsed time, not necessarily the gradient at one instant. A student who has practised straight-line gradients must keep the distinction between average and instantaneous information.
Suppose the energy transferred rises from 4 J at 1 s to 22 J at 4 s. Average power over that interval is (22 − 4) ÷ (4 − 1) = 6 W. The gradient has units J/s, which are watts. Notice that the graph did not begin at zero for the chosen interval. Dividing 22 by 4 would answer a different question and silently throw away the starting condition.
After finding a slope, do not automatically call it efficiency. A slope on an energy-time graph has units of power. Efficiency is a dimensionless input-output ratio. A graph of useful energy against total input energy might have a slope representing efficiency when the relationship and axes support that interpretation. Reading the axis labels is part of the mathematics and the physics.
What a results table can establish
A table with columns labelled trial, mass, height, lift time, electrical input and initial/final speed allows more careful interpretation than a decorative chart. Each column should contain units in the heading, not repeated ambiguously in each number. If one row has missing input energy, you may still calculate useful energy and useful power, but you cannot calculate that trial’s efficiency from the stated evidence.
If the mass changes between trials, equal heights no longer imply equal useful energy. A 2 kg load lifted 1 m gains twice as much gravitational potential energy as a 1 kg load lifted 1 m under the same gravitational field strength. If heights differ, compare equal tasks before making a blanket claim about which motor is better. If speeds differ at the endpoints, ask whether a kinetic energy correction is necessary.
The thoughtful examination answer names the comparison’s conditions. ‘Motor B had the higher measured efficiency for the same load and vertical rise in these trials’ is supported by appropriate input-output data. ‘Motor B is always better’ extends beyond the experiment. A conclusion should be neither timid nor inflated: say what the readings show, then name the condition under which they show it.
Recover when the question is worded indirectly
Questions do not always say ‘calculate power’. A prompt might ask, ‘How much useful energy does the motor transfer each second?’ That is power. It might ask, ‘What percentage of the battery energy becomes gravitational potential energy?’ That is efficiency. It might ask, ‘How much energy is transferred against gravity?’ That calls for mgh in a suitable lift. Translating the language into a quantity comes before calculation.
Underline the target but do not copy phrases unthinkingly. A battery supplying 60 J in 5 s gives average input power 12 W. That is different from useful output power if only 45 J reaches the load. A clear solution writes ‘input’ or ‘useful output’ beside power so the reader knows the boundary. Then it can calculate useful power as 9 W and efficiency as 75%.
When stuck, write the known quantities with units in a small list, label one unknown, and sketch a simple transfer arrow from input to useful output plus other transfers. This gives structure to an unfamiliar case. If the final answer has units of joules when the question asks for watts, the mismatch can be detected before handing in the script.
Common mistakes that cost easy marks
The first mistake is taking distance along a ramp as vertical height in mgh. The second is dividing energy by distance and calling the result power, forgetting that power requires time. The third is putting total input energy in the numerator of efficiency, producing a figure larger than 100%. The fourth is equating shorter time with greater energy use. None of these errors requires advanced algebra to correct; each begins with reading the meaning of the quantity.
Another common mistake is treating 65% as 65 in a calculation instead of 0.65. If the useful output is 13 J and efficiency is 65%, total input is 13 ÷ 0.65 = 20 J, not 13 ÷ 65. Students should check whether the requested value is plausibly greater than the useful output in a process that also transfers energy elsewhere.
A quieter error appears in explanations. The phrase ‘some energy was lost’ is often too vague. Say that some energy was transferred to thermal and sound pathways, where the evidence and context justify it. Not all additional transfer pathways can be assigned an exact value from a single efficiency reading. Precision means stating what is known, and refusing to label an unknown difference more specifically than the measurements allow.
A seven-day practice route
On day one, practise distinguishing energy, work, power and efficiency in plain sentences before touching a calculator. On day two, work through vertical lifts with mass, height and time. On day three, introduce electrical input energy and percentage efficiency. On day four, use the same experiment with two motors and write paired comparisons. On day five, interpret energy-time graphs and average gradients over intervals that do not start at the origin.
On day six, attempt unfamiliar motor and ramp scenarios under a small time limit. Include at least one problem where not enough data are supplied to determine efficiency. On day seven, revisit only the items that caused errors: misread axes, incorrect units, an inverted ratio or an unsupported conclusion. Record the type of error and one sentence that would prevent it next time.
This route suits an upper-secondary learner who has already met the main energy equations. A student still confusing vertical height with slope length should spend more time on the geometric foundations rather than rushing into mixed investigations. The objective is not to complete seven pages of questions. It is to reach the point where the student can explain why an equation answers a particular question and when it does not.
Practice case 1 — calculate only what is supported
A 1.5 kg load rises 0.80 m from rest to rest. Assume g = 10 N/kg. The lift takes 4.0 s, and the input meter reports 20 J. Calculate useful energy, useful power and efficiency. The useful energy is 1.5 × 10 × 0.80 = 12 J; useful power is 12 ÷ 4.0 = 3.0 W; efficiency is 12 ÷ 20 × 100 = 60%. A complete answer should include each quantity and its unit or percentage, with a single sentence identifying the rise in gravitational potential energy as the useful output.
Now remove the meter reading from the question. The useful energy and power remain calculable, but efficiency does not. Students sometimes invent a total energy figure because they think every problem must have a numerical result. The correct scientific response is that the total input energy is required and has not been supplied. Naming the missing measurement is a stronger answer than guessing.
Practice case 2 — separate operating speed from efficiency
Two motors raise identical loads to the same height. Motor X takes 2.0 s and consumes 30 J; Motor Y takes 4.0 s and consumes 24 J. Both give 18 J of gravitational potential energy. X’s useful power is 9.0 W and its efficiency is 60%. Y’s useful power is 4.5 W and its efficiency is 75%. Therefore X is more powerful for the stated useful outcome, while Y is more energy-efficient. Neither adjective is a synonym for the other.
Write a comparison that would be acceptable in a short structured response: ‘Motor X transfers the same 18 J to the load in half the time, so its average useful power is twice that of Y. Motor Y uses 24 J rather than 30 J to provide the same 18 J output, so its efficiency is higher.’ This is better than ‘X is faster but Y is greener’, because the latter introduces an environmental claim that the experiment has not established.
Practice case 3 — the interval on a graph matters
A cumulative energy graph shows 6 J at 2 s and 30 J at 8 s. Calculate average power between those readings. The change is 24 J over 6 s, so the answer is 4 W. A student who divides 30 by 8 obtains 3.75 W, which describes the average since time zero only if the graph legitimately begins with zero energy at time zero. It is not the required interval.
If a follow-up asks whether power was constant between 2 and 8 seconds, two points alone do not establish a straight line throughout. The calculated 4 W is an interval average. To claim constant power, we need an appropriate relationship or sufficient plotted points showing a straight line over that interval. The same honesty applies to a curve whose endpoints happen to suggest a convenient whole-number average.
Practice case 4 — account for speed as well as height
A 2.0 kg trolley travels upward while changing speed from 1.0 m/s to 2.0 m/s and rising 0.50 m vertically. With g = 10 N/kg, the gravitational potential energy increase is 10 J. Its kinetic energy changes from 1 J to 4 J, an increase of 3 J. The total increase in these two stores is therefore 13 J. If the purpose is to provide both height and speed, the useful outcome may include both terms.
If the total supplied energy is 20 J and the question explicitly defines useful output as those two increases, efficiency is 13 ÷ 20 = 65%. If it defines useful output solely as the gain in gravitational potential energy, the efficiency for that narrower task is 50%. This case shows why ‘efficiency’ is not a number that can be extracted without reading what output the problem counts as useful.
Practice case 5 — repair an inverted ratio
A learner writes: ‘Input = 27 J, useful output = 18 J, so efficiency is 27/18 × 100% = 150%.’ Identify the error. The numerator and denominator have been reversed. For the defined process, efficiency is useful output divided by total input: 18/27 × 100% = 66.7% to three significant figures. The correction changes the meaning of the ratio, not just the arithmetic.
A complete response can add that an efficiency above 100% for a conventional energy-transfer device contradicts the defined accounting because the useful output cannot exceed the total energy input. Be cautious about saying every measured percentage over 100% proves fraud: experiments can include reading errors, uncounted input stores or a mistaken boundary. The correct evaluation is to investigate the data and assumptions.
Practice case 6 — compare rated and measured quantities
A device has ’50 W’ printed on its casing. It lifts a load through a task lasting 3 s. Can we conclude that exactly 150 J entered the device? Only if the stated 50 W represents average input power during that particular interval. A rating alone may describe nominal or maximum operating conditions. The label is not automatically a measurement of this trial’s input.
If an independent meter instead records a steady average input of 46 W over those 3 s, input energy is 138 J. A written solution should cite the measured 46 W and the duration rather than the 50 W label. This distinction is essential when a question includes information that looks authoritative but is not the appropriate observation.
Practice case 7 — make a fair test
A class compares motors but changes the load mass, lift height, timing method and battery condition between trials. The resulting efficiencies differ. The question asks which motor was more efficient. You should first note that the trials were not directly comparable. A fairer investigation uses equivalent loads, rises, motion endpoints and measurement methods, while recording the actual energy input for each.
Repeat trials to assess scatter. Randomise or alternate motor order if heating or battery depletion might influence later measurements. Report which factors remain uncontrolled. A short school practical may not permit perfect isolation, but it should not award a broad claim from a comparison with obvious confounding differences.
Practice case 8 — read a conclusion before accepting it
A poster states, ‘Motor A is twice as good because its power is 10 W and B’s power is 5 W.’ Ask what ‘good’ means. If the relevant objective is lifting quickly under a fixed useful task, greater useful power may be advantageous. If the objective is minimising energy use, the conclusion needs input and efficiency measurements. A twofold difference in power does not by itself show a twofold difference in energy efficiency.
A model response distinguishes a measured performance variable from an overall judgement: ‘Motor A has twice the reported power, but the poster provides no total energy input, so energy efficiency cannot be compared.’ That sentence does not dismiss the power data. It puts the data in the proper argumentative place.
Practice case 9 — choose the right denominator
A motor delivers 16 J to the raised load and an additional 4 J to other useful mechanical output as explicitly defined by the task, while total input is 32 J. If useful output includes both changes, efficiency is (16 + 4) ÷ 32 = 62.5%. If the task is exclusively lifting and the other mechanical effect is not useful, the corresponding lift efficiency is 16 ÷ 32 = 50%. The same input yields different task-defined percentages.
Do not add an output twice. If the problem already gives total useful mechanical energy as 20 J, adding the 16 J gravitational part again would double-count. State your definition once, then use the terms that belong to it. This is the energy version of avoiding a double-counted subtotal in a mathematics table.
Practice case 10 — inverse calculation from efficiency
A lifting device has a useful output of 36 J and a measured energy efficiency of 72%. Find the input energy. Use 0.72 = 36 ÷ input, so input = 36 ÷ 0.72 = 50 J. The other-transfer allowance in this simple account is 14 J. Notice that solving for a denominator requires rearranging before substituting; multiplying 36 by 0.72 would incorrectly make the input smaller than its useful output.
Add a second question: how long would a steady average input power of 25 W need to supply 50 J? Time = energy ÷ power = 50 ÷ 25 = 2.0 s. This is a combined problem, but each step is ordinary once the quantities have been named. The student should finish by checking that joules divided by watts yield seconds.
Practice case 11 — write a data-based explanation
A student sees three trials of the same lift with input energies 25 J, 23 J and 22 J, but the sheet only records one useful-output figure of 18 J without specifying whether it applies to every trial. The sound answer asks whether mass, height and endpoint motion were held constant. If the useful outcome was identical, the corresponding efficiencies are 72%, about 78.3% and about 81.8%. If it was not, the comparison is incomplete.
The data alone also do not explain why input energy changed. Friction, starting conditions, battery state, measurement uncertainty or changes in motor operation are possible factors, not established causes. Use conditional phrasing when evidence is conditional. An examiner can reward a clear distinction between a measured trend and a proposed explanation.
Practice case 12 — an unfamiliar context
A small model crane raises a package. The question gives a 12 N upward lifting force, a 0.40 m upward displacement and an elapsed time of 1.5 s. If the force is aligned with the movement and the simplified work model applies, work done by the stated force is 12 × 0.40 = 4.8 J; average power associated with that work is 4.8 ÷ 1.5 = 3.2 W. The force value has a different job from an input-energy meter reading.
Suppose the question next reports 6.4 J supplied. If 4.8 J is designated useful, efficiency is 4.8 ÷ 6.4 = 75%. Explain the model assumptions rather than silently treating every measured force as a motor’s total input. This final drill tests transfer: the apparatus changed from a trolley ramp to a crane, but the underlying distinction stayed intact.
How to answer a structured examination question well
Start with a short statement of the physical relationship. Substitute quantities in SI units. Show the main operation. Report the answer with a sensible number of significant figures, and write the correct unit or percentage. If the question asks to explain or compare, add the causal or quantitative statement in words. For example: ‘Both motors raise the same load to the same height; the first takes less time, so its average useful power is greater.’
Do not let a long paragraph replace a required calculation, and do not let a bare number replace an explanation. The command word matters. ‘State’ can require a short fact. ‘Calculate’ calls for a numerical route. ‘Explain’ needs a mechanism or a reason. ‘Evaluate’ may need a judgement about the reliability and limits of an inference. Match the requested form before expanding.
If time is running short, protect the high-certainty marks: write the correct formula, substitute known quantities, show any rearrangement and give units. If a later part requires interpretation, use the number already obtained as evidence. Avoid copying the question into the answer. Clear, short sentences are often enough when the thinking is sound.
Frequently asked questions
Is a 100 W motor always better than a 50 W motor? Not necessarily. A power rating concerns a rate under stated conditions, not the total useful output, efficiency, reliability or suitability for every task. Compare the actual required duty and measured input and output.
Can a motor be efficient and slow? Yes. Efficiency and power answer different questions. A motor can convert a high proportion of input energy into the defined useful output while taking longer to complete the transfer.
Does friction destroy energy? No. Friction transfers energy into other stores, often increasing internal energy. It may reduce the share available for the intended useful motion.
Is input power the same as useful power? Only in an idealised 100% efficient process under comparable time conditions. Real devices often transfer energy elsewhere, so label each quantity.
Should I use a motor label or the trial measurement? Use the quantity the question defines and the evidence it supplies. A rating does not automatically represent actual consumption in a particular interval.
Where should I revise if the algebra is weak? Start with ratios and rearranging equations in the Mathematics Hub. Stronger arithmetic supports clearer physics; it does not replace understanding of energy boundaries.
What to read next
Return to Vol 0683: trolley-ramp energy transfers if you need the physical stores, and Vol 0684: speed-squared graph analysis if gradients or model assumptions are fragile. Use the Science Hub for evidence and systems, Mathematics Hub for ratios and equations, and Examination Craft for performing reliably under timed conditions.
For students making the PSLE-to-secondary transition, the Primary 6 transition guide and PSLE Science guide provide an earlier route into observation and evidence. These are links for learning continuity, not claims that a PSLE formula and a G3 Physics assessment are identical. The definitive examination specification remains SEAB’s K323 document.
