S4 Physics: Work, Energy and Power — Tutor Lesson Plan (60 min, W = Fs to efficiency, four errors)

2026-11-02

Tutor lesson-prep guides · Secondary physics (DSE)

Students multiply force by distance in every direction and lose the marks that separate a grade 4 from a grade 5. This lesson fixes the angle first, then lets the numbers do the persuading: 3924 joules, 784.8 watts, 83.1 per cent.

Objectives — the student can decide with W = Fs cos(theta) whether a force does work and name the three zero-work cases, convert between kinetic and potential energy and apply conservation to a fall, a ramp and a braking car, and use P = W/t with an efficiency ratio to compare the same task done at different speeds.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Weight and mass	What is the weight of 50 kg?	490.5 newtons
Speed and displacement	How do speed, distance and time relate?	Speed is distance divided by time
Ratio arithmetic	Twice the time at the same power means what?	Twice the energy transferred

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-8 min	Hook	A suitcase carried across a hall	Say how much work the hand does
8-20 min	Scene 1	Three conditions, three zero-work cases	Classify four examples
20-32 min	Scene 2	Kinetic and potential energy exchange	Get 3924 joules and 12.5 metres per second
32-44 min	Scene 3	Power: same work, different time	Get 784.8 watts
44-54 min	Scene 4	Efficiency on a ramp	Get 83.1 per cent
54-60 min	Exit check	Three items in section 08	Marked at once

03 Teaching script (read it out)

Scene 1: Three conditions for work

The definition W = Fs cos(theta) only counts the part of the displacement along the force. Carry a suitcase 15 metres across a hall while pushing upwards and the force is perpendicular to the displacement, so cos 90 degrees is zero and the work done by that force is zero. Your muscles still burn energy, but that is biology, not W = Fs. The three zero cases are worth naming aloud: force perpendicular to motion, no displacement at all such as a wall that will not move, and no force at all such as a curling stone sliding at constant speed. Friction on the same 15 metres with 40 newtons gives minus 600 joules, and the minus sign is not decoration: it is energy leaving the object as heat and sound.

Ask: “If the hand does no work on the suitcase, who did?” Nobody, in the physics sense, and that answer is the whole point of the first eight minutes.

Scene 2: Energy only changes form

Lift 50 kg to 8 metres and the potential energy gained is mgh, which is 50 times 9.81 times 8, so 3924 joules. Let it fall and the kinetic energy just before landing equals 3924 joules, so v is the square root of 2gh, about 12.5 metres per second, and the mass has cancelled. That cancellation is worth showing algebraically because students believe heavy objects fall faster. For the car, 1200 kg at 20 metres per second holds 240 kilojoules, so stopping over 40 metres needs an average braking force of 240000 divided by 40, which is 6000 newtons.

Warn: doubling the speed quadruples both the kinetic energy and the braking distance. That is the physics behind a speed limit, and it is a better argument than telling anyone to be careful.

Scene 3: Power is work per second

The work of lifting 50 kg by 8 metres is fixed at 3924 joules. Done in 5 seconds it is 784.8 watts; done in 20 seconds it is 196.2 watts. The work never changed, only the rate, and that distinction is what P = W/t encodes. On a machine the same statement is P = Fv, useful when a force moves at a steady speed. Efficiency compares what you get with what you paid: if 5000 joules go in and 3924 joules become potential energy, the efficiency is 78.5 per cent and the missing 1076 joules have become heat. On a 20-metre ramp the ideal force is 3924 divided by 20, so 196.2 newtons; add 40 newtons of friction and the real force is 236.2 newtons, the input work 4724 joules and the efficiency 83.1 per cent. A ramp saves force, never energy, and students should say that with their own numbers.

Ask: “Why is the ramp more efficient than lifting straight up?” Because friction costs less than the losses of a direct lift in that comparison, and both figures must be computed the same way.

04 Question types and worked steps

Type	Example	Steps	Answer
Calculation	Find the work done lifting 50 kg by 8 metres.	W = mgh = 50 times 9.81 times 8	3924 joules
Calculation	A 1200 kg car at 20 metres per second brakes to rest in 40 metres. Find the force.	Half mv squared, then divide by distance	6000 newtons
Calculation	Find the power of the lift when it takes 5 seconds.	P = W over t = 3924 over 5	784.8 watts
Explanation	Find the efficiency of a 20-metre ramp with 40 newtons of friction.	Useful work over input work	83.1 per cent

Sanity check (teach this): Every answer must carry its unit and the direction test: watts for power, joules for energy, newtons for force, and a percentage that cannot exceed a hundred. A braking answer is checked by multiplying force by distance back into the kinetic energy, and a power answer by multiplying back by time. If a student writes 3924 joules as a power, the number is right and the concept is wrong, which is exactly the error this lesson exists to catch.

05 Independent practice (3 questions)

Question	Answer	What to watch
A 50 kg student runs up stairs 8 metres high in 5 seconds. Find the power.	784.8 watts.	Check for a missing g
A 1200 kg car at 20 metres per second stops with 6000 newtons of braking force. Find the distance.	40 metres.	Work backwards from the energy
A 20-metre ramp rises 8 metres with 40 newtons of friction. Find the efficiency.	83.1 per cent.	Define useful and input first
Explain why carrying a box horizontally does no work on the box.	The force is vertical and the displacement horizontal, so the angle is 90 degrees.	Name the angle, not the effort

06 Common mistakes and what to say

Mistake	Student writes	Say this
Assuming force times distance always counts	Carrying a case horizontally and using W = Fs	The force and the displacement must share a direction; perpendicular means zero work, and your 5 metres do not help
Putting weight into the kinetic energy formula	Half times 500 times 20 squared, giving 100 kilojoules	Kilograms and newtons are not interchangeable; 500 newtons is about 51 kilograms
Confusing power with energy	Answering that the power is 3924 joules	A joule is energy, a watt is joules per second, and the wrong unit is a wrong concept
Reporting an efficiency above 100 per cent	Answering 120 per cent	Above a hundred means useful and input work are swapped, or friction was left out

07 Homework (paper, section and question numbers)

Secondary physics chapter index — Read this chapter with the forces and the momentum chapters
HKDSE Physics Paper 1 practice paper — Work and energy questions; redo section 05 questions 1 to 4 of this guide

08 Exit check (last 5 minutes)

Question	Answer	Decision
Name the three cases where a force does no work.	Perpendicular force, no displacement, no force.	All correct: move to momentum
How much work lifts 50 kg by 8 metres?	3924 joules.	Wrong: redo mgh
What happens to the braking distance when speed doubles?	It quadruples.	Two or more wrong: back to scene one

09 Tutor FAQ

Do students need cos(theta) at DSE?

They need the idea, not the trigonometry. Saying the force must share the direction of motion carries the mark.

Should they memorise 9.81?

Yes, and they should keep the units attached to every line, because unit slips cost more marks than arithmetic slips in this topic.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.