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S4 Physics | Work, Energy and Power prep guide (for tutors) | W = Fs, KE, PE and power
Students multiply force by distance every time, so carrying a 10 kg box across the room counts as work. This lesson fixes one rule first: only displacement along the force counts.
Objectives — Students can: (1) apply W = Fs cosθ and name the three zero-work cases; (2) solve slope, falling and braking problems with energy conservation; (3) compare the same job done over different times using P = W/t and efficiency.
01 Prerequisites (3-minute check)
If any of these fails, fix it before the lesson starts.
| Skill | Quick check | Pass mark |
|---|---|---|
| Weight and mass | weight of 50 kg | 490.5 N |
| Speed and displacement | v and s in uniform motion | v = s / t |
| Ratio reasoning | twice the time, same work | half the power |
02 Lesson flow (60 minutes)
| Time | Stage | Tutor does | Student does |
|---|---|---|---|
| 0-10 min | When work is done | W = Fs, force along displacement | name the zero cases |
| 10-24 min | Lifting and power | 50 kg raised 8 m | 3924 J, 784.8 W |
| 24-36 min | Kinetic energy and braking | 1200 kg at 20 m/s | 240 kJ to 6000 N |
| 36-46 min | Slopes and efficiency | 20 m ramp, 40 N friction | 83.1% |
| 46-56 min | Conservation mixed problems | free fall and springs | write the equation first |
| 56-60 min | Exit check | three questions, marked at once | independent |
03 Teaching script (read it out)
Scene 1: Work needs force, displacement, and no right angle between them
W = Fs cosθ only counts the component of displacement along the force. Lift a case and walk {:.0f} m: the force is upward, the displacement horizontal, cos 90° = 0, so the work done by your hand is zero. Muscles still burn chemical energy, but that is biology, not W = Fs.
Three zero-work cases: force perpendicular to motion (carrying a case, centripetal force); no displacement (pushing a wall); no force (a puck gliding on ice).
Positive and negative work: friction opposes motion and does -600 J (40 N x 15 m). Negative work means energy is taken away, not destroyed: it becomes heat and sound.
Three zero-work cases: force perpendicular to motion (carrying a case, centripetal force); no displacement (pushing a wall); no force (a puck gliding on ice).
Positive and negative work: friction opposes motion and does -600 J (40 N x 15 m). Negative work means energy is taken away, not destroyed: it becomes heat and sound.
Task: decide which of four situations does zero work (carrying a case, pushing a wall, a lift rising, a ball on a string), and give one reason each.
Scene 2: Energy only changes form: kinetic and potential
Raise 50 kg by 8 m: gain in potential energy = mgh = 50 x 9.81 x 8 = 3924 J. Release it and just before landing the kinetic energy is 3924 J, so v = sqrt(2gh) = 12.5 m/s, independent of mass.
Car: 1200 kg at 20 m/s has KE = 0.5 x 1200 x 20^2 = 240000 J = 240 kJ. To stop within 40 m the average braking force is 240000 / 40 = 6000 N.
Double the speed and the kinetic energy, and the stopping distance, both quadruple. That argument beats “drive carefully”.
Car: 1200 kg at 20 m/s has KE = 0.5 x 1200 x 20^2 = 240000 J = 240 kJ. To stop within 40 m the average braking force is 240000 / 40 = 6000 N.
Double the speed and the kinetic energy, and the stopping distance, both quadruple. That argument beats “drive carefully”.
Task: the same car goes from 20 m/s to 40 m/s. How many times larger are KE and stopping distance? (4 and 4) Explain why not 2.
Scene 3: Power and efficiency: same work, different speed
Raising 50 kg by 8 m is always 3924 J of work. Done in 5 s: P = 3924 / 5 = 784.8 W. Done in 20 s: 196.2 W. Same work, 4 times the power.
Efficiency: input 5000 J, useful output 3924 J, so efficiency = 78.5%; the other 1076 J leaves as heat.
Ramp: ideal force 196.2 N, plus 40 N friction, so 236.2 N applied over 20 m, input 4724 J, efficiency 83.1%. A ramp saves force, never work; make students say that with their own numbers.
Efficiency: input 5000 J, useful output 3924 J, so efficiency = 78.5%; the other 1076 J leaves as heat.
Ramp: ideal force 196.2 N, plus 40 N friction, so 236.2 N applied over 20 m, input 4724 J, efficiency 83.1%. A ramp saves force, never work; make students say that with their own numbers.
Task: compute the power for lifting 50 kg by 8 m in 5 s (784.8 W), then say whether a pulley that halves the force changes the power.
04 Question types and worked steps
| Type | Example | Steps | Answer |
|---|---|---|---|
| Calculation | W = Fs | check force and displacement are aligned | 3924 J |
| Calculation | KE and braking | 0.5mv^2 then divide by distance | 6000 N |
| Calculation | Power | P = W/t or Fv | 784.8 W |
| Explanation | Efficiency | useful output over input | 83.1% |
Sanity check (teach this): Units: work and energy in J (N m), power in W (J/s); kinetic energy needs kg and m/s. Students mix g = 10 and 9.81 within one question, so fix it: this page uses 9.81.
Order of magnitude: a 1200 kg car at 20 m/s carries hundreds of kJ, not hundreds of J. A braking force of a few newtons means a missing half or the wrong mass.
Signs: negative work removes energy; if total energy rises, an input was missed.
Order of magnitude: a 1200 kg car at 20 m/s carries hundreds of kJ, not hundreds of J. A braking force of a few newtons means a missing half or the wrong mass.
Signs: negative work removes energy; if total energy rises, an input was missed.
05 Independent practice (3 questions)
| Question | Answer | What to watch |
|---|---|---|
| A 50 kg student runs up 8 m of stairs in 5 s. Power? | 784.8 W | missing half or g |
| A 1200 kg car stops from 20 m/s with 6000 N. Distance? | 40 m | can they reverse it |
| A 20 m ramp lifts 8 m with 40 N friction. Efficiency? | 83.1% | definitions of useful and input work |
06 Common mistakes and what to say
| Mistake | Student writes | Say this |
|---|---|---|
| Assuming any force times distance is work | W = Fs while carrying a case | Your hand pushes up, the case moves sideways: 90 degrees, so zero work. |
| Using weight as mass in KE | 0.5 x 500 x 20^2 | Newtons and kilograms are not interchangeable; 500 N is about 51 kg. |
| Confusing power with energy | answering 3,924 W | Joules are energy, watts are joules per second; a wrong unit is a wrong idea. |
| Efficiency above 100% | answering 120% | Over 100% means useful and input work are swapped, or friction was ignored. |
07 Homework (paper, section and question numbers)
- Secondary Physics chapter index — all chapters
- DSE Physics practice paper — work and energy questions 5-11
08 Exit check (last 5 minutes)
| Question | Answer | Decision |
|---|---|---|
| Three zero-work cases | perpendicular, no motion, no force | all correct: move on to momentum |
| Work to raise 50 kg by 8 m | 3924 J | wrong: reteach mgh |
| Stopping distance when speed doubles | four times | two or more wrong: back to act one |
09 Tutor FAQ
Why is carrying a case not work?
The force is upward and the displacement horizontal, so cos 90° = 0. The body still spends chemical energy, but W = Fs counts mechanical transfer only.
Why does landing speed not depend on mass?
mgh = 0.5mv^2 cancels m on both sides, leaving v = sqrt(2gh).
Does a ramp save work?
It saves force but not work: a smaller force acts over a longer distance. With friction the input work is larger than lifting straight up.
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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.
