Work & Power · Grade 12 Science
Chapter: Work, Energy & MomentumWork & Power
Learning goals
- I can calculate the work done by a force using W = Fd (and W = Fd·cosθ when the force acts at an angle).
- I can apply the work-energy theorem and calculate power with P = W/t.
Readings for this unit
Before this unit
Force components, displacement, and time.
Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.
Start with something to explain
Why can equal lifting jobs require different motor power?
Illustrative evidence case. Two motors raise identical loads to the same height in different times.
Notice and wonder
Record two observations using the stated quantities or relationships. Keep observations separate from possible causes. Write one question that a new measurement could help answer.
Why can equal lifting jobs require different motor power?
Your first model
Record force, height, time and the selected useful output.
Build the language from the idea
- work
- The energy transferred to an object when a force acts on it through a displacement: W = Fd·cosθ, measured in joules (J).
- joule (J)
- The SI unit of work and energy; one joule is one newton-metre (1 J = 1 N·m).
- work-energy theorem
- The net work done on an object equals its change in kinetic energy: W_net = ΔKE.
- power
- The rate at which work is done or energy is transferred: P = W/t, measured in watts (W).
- watt (W)
- The SI unit of power; one watt is one joule per second (1 W = 1 J/s).
In everyday speech 'work' means almost any effort, but in physics it has a sharp, specific meaning: work is the energy a force transfers to an object as that object moves. When a force pushes something through a displacement in the same direction, the work done is simply W = Fd — force times distance. Because a joule is a newton-metre, pushing with 10 N through 3 m transfers 30 J of energy. Work is the bridge between force (Unit 2) and energy: a force by itself does nothing to an object's energy until the object actually moves.
Direction matters, which is why the full formula is W = Fd·cosθ, where θ is the angle between the force and the displacement. When you carry a heavy backpack horizontally across a room, gravity pulls straight down while you move sideways — the angle is 90°, cos 90° = 0, and gravity does zero work on the pack even though holding it feels exhausting. When friction drags backward against motion, the angle is 180°, cos 180° = −1, and friction does negative work, removing energy from the object. Only the part of a force that lies along the motion actually transfers energy.
The payoff of tracking work is the work-energy theorem: the *net* work done on an object equals its change in kinetic energy, W_net = ΔKE. Push a cart and the positive net work speeds it up; let friction act and the negative net work slows it down. This theorem is a shortcut of real power — if you know the net work, you know exactly how much the object's motion energy changed, without having to follow the messy details of acceleration and time in between.
Doing work is one thing; how *fast* you do it is another, and that is power. Power is the rate of doing work, P = W/t, measured in watts (one watt is one joule per second). Two cranes might lift identical loads to the same rooftop — doing exactly the same amount of work — yet the crane that finishes in half the time delivers twice the power. Power can also be written P = Fv, the product of force and speed, which is why a car engine must supply more power to hold a high speed against air resistance than to cruise slowly.
These ideas quietly rate almost every machine around you. The 'horsepower' on an engine, the wattage on a motor or a light bulb, the output of a power plant — all are statements about how quickly energy is being transferred. Engineers care about both totals and rates: a small motor and a huge one might eventually do the same job, but only the powerful one can do it fast enough to matter. Master work and power and you hold the vocabulary for describing every energy transfer in the rest of this unit.
Model check: A force perpendicular to displacement does no mechanical work in the point-particle model. W=30 J; P=W/t=15W, if that force is constant.
Build your understanding
Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.
Part 1 · Define work by force and displacement
Work is energy transferred by a force through displacement. For a constant force, W=FΔx cosθ uses the angle between the vectors. Positive work adds energy through that interaction; negative work removes it. A perpendicular force does zero work in this model even though it can turn the motion. Specify which force’s work you calculate and distinguish it from the net work of all forces.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
A 10 N force acts perpendicular to a 3 m displacement. What work does it do?
Visual reasoning — after your attempt
W=10×3×cos 90°=0 J.
For a constant force, work is Fd cosθ, with θ between force and displacement. Work describes energy transfer, not effort or elapsed time alone. A perpendicular force does no work in this ideal account even while it changes motion direction.
Holding a stationary load can require bodily energy while doing zero mechanical work on that load through displacement. The system and transfer pathway matter.
Pause and explain
Why can a force do zero work while present?
Check your reasoning after answering
The object may have zero displacement or the force may be perpendicular to displacement, giving a zero dot product.
Part 2 · Use a force-position area when force varies
If force changes with position, multiply-and-add small contributions or find the signed area under its force–position graph. For a linear spring stretched slowly from x=0 to 0.4 m with k=50 N/m, applied force rises from 0 to 20 N. The triangular area is 4 J. Multiplying the final 20 N by the whole extension gives twice the correct work because that final force did not act over the entire interval.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
What average applied force produces the same 4 J over 0.4 m?
Visual reasoning — after your attempt
Favg=W/Δx=10 N, the average of 0 and 20 N for this linear ramp.
For one-dimensional motion, signed area under a force-versus-position curve gives work. A force-versus-time area gives impulse instead. Match axes and units before using geometric areas. Negative work can reduce the selected object’s kinetic energy when no other work offsets it.
A linear force increasing from 0 to 20 N over 0.50 m has triangular work area 5.0 J.
Pause and explain
Why is force-time area not work?
Check your reasoning after answering
Its units are N·s and it accounts for momentum change; work requires displacement and has units N·m or joules.
Part 3 · Distinguish power from total transferred energy
Power is energy transferred per time: average P=W/Δt for this process. A watt is one joule per second. Two machines can perform the same work while having different power because one takes less time. A power rating alone does not give total energy without duration and actual operation. Distinguish average power over a stated interval from brief peak power and keep seconds consistent when computing watts.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
How much work does a steady 100 W device transfer in 12 s?
Visual reasoning — after your attempt
W=PΔt=1200 J, assuming the stated steady output applies for all 12 s.
Average power is energy transferred divided by elapsed time, commonly W/Δt for work. A larger power can transfer the same energy in less time. For a force parallel to velocity, instantaneous mechanical power can be Fv; direction matters in the general dot product.
Two motors can lift the same load to the same height with the same ideal work while taking different times and delivering different average power.
Pause and explain
Does greater power necessarily mean greater total work?
Check your reasoning after answering
No. The duration also matters; power describes a transfer rate, while work describes the transferred amount.
Part 4 · Connect output and input with efficiency
Efficiency compares useful output with total input for a defined process and service. If 1000 J enters and 350 J is delivered as the selected useful output, efficiency is 35%; the remaining 650 J must be accounted for as other transfers or storage. Energy is not destroyed by inefficiency. What counts as useful can differ between a lamp and a heater, so comparisons require the same service and a consistent system boundary.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
Does 35% efficiency mean 650 J disappeared?
Visual reasoning — after your attempt
No. It left through other transfers or remained stored within the defined accounting interval.
Efficiency compares useful output to input for a stated job and interval. Mechanical output is not automatically equal to electrical input; other outputs and changes in stores must be accounted for. Define the useful service before reporting a percentage.
An 80%-efficient lift needing 100 J of useful output requires 125 J input in that stated steady account.
Pause and explain
Why must efficiency name the useful output?
Check your reasoning after answering
Different useful services and boundaries can produce different ratios. A percentage without the job or interval is ambiguous.
Apply the whole unit · Compare lifting work and motor input
Illustrative motor lifts a 20.0 kg load vertically by 3.00 m in 5.00 s at constant speed. Use g=10.0 m/s² and overall efficiency 0.750.
Your task
Find ideal useful work, output power, input energy, and average input power. Explain why constant speed does not mean the motor transfers no energy.
What your response should show
- Define the system, quantities, units, and relevant conditions before selecting a model.
- Show intermediate calculations or a labeled evidence account, including signs or denominators.
- Explain how the supplied evidence supports the conclusion rather than reporting only an answer.
- Identify a meaningful uncertainty, model limit, or measurement needed to test the conclusion.
Response guidance — after your attempt
Useful work=mgh=20×10×3=600 J.
Output power=600/5=120 W.
Input energy=600/0.750=800 J; average input power=160 W.
The motor’s upward force does positive work while gravity does negative work, so load kinetic energy can remain constant. The useful increase is gravitational potential energy of the load-Earth system. The remaining 200 J enters other outputs or stores within the assumed efficiency account; it is not destroyed.
Scientific references and source-model foundations
Parent / teacher teaching notes
Ask students to name the system for potential energy and the interval for power. Check that input exceeds useful output for the supplied efficiency. Ask the learner to explain each symbol, axis, unit and arrow before calculating. Calculated examples, supplied illustrative records and actual historical observations are labeled separately. Selected public OpenSciEd middle-school source figures are foundation examples, not complete high-school editions or an endorsement; all advanced explanations are supplied locally.
The key idea
Work is energy transferred by a force acting through a displacement (W = Fd·cosθ); the net work on an object changes its kinetic energy; power (P = W/t) is how fast that work is done.
Work
W = Fd·cosθ
F is the force in newtons (N); d is the displacement in metres; θ is the angle between the force and the displacement; W is in joules (J). When the force is along the motion, cosθ = 1 and W = Fd.
Work-Energy Theorem
W_net = ΔKE = ½mv_f² − ½mv_i²
the net work done on an object equals its change in kinetic energy; v_i and v_f are the initial and final speeds.
Power
P = W / t (also P = Fv)
W is work in joules; t is time in seconds; P is power in watts (1 W = 1 J/s); v is speed in m/s.
Worked examples
Example 1
- Only the part of the force along the motion does work, so use W = Fd·cosθ.
- W = 50 × 4.0 × cos 60°.
- cos 60° = 0.5, so W = 200 × 0.5 = 100.
Example 2 · Worked application
- Only the aligned force component contributes: W=Fd cosθ.
- W=40×5×0.5=100 J.
- P_avg=W/t=100/4=25 W.
State which force's work is calculated.
Example 3
- Power is the rate of doing work: P = W/t.
- P = 9,000 ÷ 6.0.
- = 1,500.
Example 4 · Find and fix the mistake
- Test the claim against the stated conditions: State which force's work is calculated.
- Power is energy transfer per time, measured in watts; work/energy is in joules.
State which force's work is calculated.
Example 5 · Reason from the visual model
- Power is energy transferred per time: average P=W/Δt for this process. A watt is one joule per second. Two machines can perform the same work while having different power because one takes less time. A power rating alone does not give total energy without duration and actual operation. Distinguish average power over a stated interval from brief peak power and keep seconds consistent when computing watts.
- W=PΔt=1200 J, assuming the stated steady output applies for all 12 s.
Illustrative equal-work comparison; durations are supplied, not inferred from bar widths.
Avoid these mistakes
- A force perpendicular to displacement does no mechanical work in the point-particle model.
Check your understanding
A 10N parallel force moves a box 3 m in 2 s. Find work and average power.
Discuss the answer
W=30 J; P=W/t=15W, if that force is constant.
Return to the phenomenon
Return to the initial case. Why can equal lifting jobs require different motor power? Revise your drawing or explanation using two named ideas and one stated limitation.
Compare explanations after your attempt
Work depends on the force–displacement account; power divides transfer by time, and input exceeds useful output for an inefficient system.
What should we investigate next?
Which comparison would test a remaining uncertainty? State what you would change, keep comparable and measure.
Investigate, calculate and revise
Keep a first model and an evidence-supported revision. Attempt each task before opening guidance; all required explanations and supplied-data routes are available locally.
Activity 1 · Return to the opening case
Use the opening phenomenon, unit definitions and explained Parts 1–4.
Your task
Return to the initial case. Why can equal lifting jobs require different motor power? Revise your drawing or explanation using two named ideas and one stated limitation.
Success criteria
- Use the stated observations and appropriate units.
- Connect a labeled model with the physical/environmental mechanism.
- State assumptions and an evidence limitation.
- Explain a substantive revision or next test.
Hint
Keep object/material identity, material transfer and energy account distinct.
Response guidance — discuss after your attempt
Work depends on the force–displacement account; power divides transfer by time, and input exceeds useful output for an inefficient system.
Activity 2 · Transfer to a changed case
Cover the worked answer and use the explained models, relationships and notebook.
Your task
A motor raises a 20 kg load by 3 m in 6 s at 70% efficiency. Find useful work, output power and input energy.
Success criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Hint
Name known quantities and conditions, show intermediate steps and check conservation.
Response guidance — discuss after your attempt
Useful work=mgh=588 J. Output power=98 W. Input energy=588/0.70=840 J; other transfers/storage total 252 J under the stated boundary.
Words to know
- work
- The energy transferred to an object when a force acts on it through a displacement: W = Fd·cosθ, measured in joules (J).
- joule (J)
- The SI unit of work and energy; one joule is one newton-metre (1 J = 1 N·m).
- work-energy theorem
- The net work done on an object equals its change in kinetic energy: W_net = ΔKE.
- power
- The rate at which work is done or energy is transferred: P = W/t, measured in watts (W).
- watt (W)
- The SI unit of power; one watt is one joule per second (1 W = 1 J/s).
James Watt and the invention of 'horsepower'
When Scottish engineer James Watt was selling his improved steam engines in the 1770 s, customers wanted to know how many horses each machine could replace. So Watt measured how much work a strong draft horse could do — hauling loads out of a mine — and defined one horsepower as about 550 foot-pounds per second, which works out to roughly 746 watts. It was brilliant marketing dressed as physics, and it stuck: car engines are still rated in horsepower today, even though the SI unit of power, the watt, is named after the very same man.
Misconception: holding a heavy weight is 'hard work' but does no physics work
Stand holding a heavy box perfectly still and your arms will soon burn — yet in the language of physics you are doing zero work on the box. Work is W = Fd, and if the box does not move, d = 0, so no energy is transferred to it. Why does it feel so tiring? Your muscle fibres are rapidly twitching and relaxing to hold the tension, doing internal biological work and generating heat inside you, but none of that energy goes into the box. It is a sharp reminder that the physics definition of work is about energy transferred to the object, not about human effort.
Negative work and the angle in W = Fd·cosθ
The cosine in W = Fd·cosθ is not a technicality — it decides the sign of the work. When a force points partly against the motion, cos θ is negative and the force removes energy: friction sliding a box to a stop (θ = 180°, cos = −1) does negative work, draining kinetic energy into heat. When a force is exactly perpendicular to the motion (θ = 90°, cos = 0), it does no work at all — which is why the tension in a string does zero work on a ball whirling in a circle, and why gravity does no work on a satellite in a circular orbit even though it is always pulling. Only the component of force along the displacement transfers energy.
Real-world power: from a light bulb to a rocket
Power ratings span an astonishing range. A human can sustain about 100 W of useful mechanical output for hours (roughly the power of an old light bulb), while a trained sprinter can hit over 1,000 W for a few explosive seconds. A family car's engine peaks near 100,000 W (about 130 hp), and the first stage of a Saturn V rocket delivered on the order of 10¹¹ W at liftoff. Since all of these do work by transferring energy, they can be compared on a single scale of watts — the same unit whether you are rating a nightlight or a moon rocket.
In the real world
Engineers rate every motor, engine, and power plant by its power — how fast it can transfer energy — because a job done too slowly (charging a car, pumping floodwater, lifting an elevator) can be as useless as not doing it at all.
Exit check
A10N parallel force moves a box 3 m in 2 s. Find work and average power.
Check your explanation after answering
W=30 J; P=W/t=15W, if that force is constant.
Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.