The shape of the lesson
By the end of the lesson, learners can
- Two kinds of energy: In the lab, release the ball on the slope with friction set to zero and watch the bar chart: potential energy drains into kinetic energy and the total bar stays the same height.
- The work-energy principle: When friction, resistance or a driving force does work, keep an energy account.
- Driving uphill: With a driving force, its work goes in on the left of the account.
- KE \(= \tfrac{1}{2}mv^2\) at the start and the end.
- PE change \(= mg \times\) vertical height change (not distance along a slope).
- Work in: driving forces. Work out: friction and resistance (\(\mu R \times d\)).
- Equation: initial KE \(+\) PE \(+\) work in \(=\) final KE \(+\) PE \(+\) work out.
- Curved paths: energy is the only method you need.
One picture every idea returns to
\(\text{KE} = \tfrac{1}{2}mv^2\) and gravitational \(\text{PE} = mgh\). If only gravity does work, \(\text{KE} + \text{PE}\) is constant. Otherwise use the work-energy principle: initial energy \(+\) work done by driving forces \(=\) final energy \(+\) work done against resistances.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Energy in, energy out.. Kinetic and potential energy turn into each other as a particle moves up and down. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Two kinds of energy | In the lab, release the ball on the slope with friction set to zero and watch the bar chart: potential energy drains into kinetic energy and the total bar stays the same height. Project tab 2 of the lab (Work, energy and power lab, kinetic and potential energy) and let a student drive it while the class predicts. Model the worked example A smooth curved slide (Fig. 1) on the board, one line at a time. Your turn: A particle slides from rest down a smooth curved track and reaches the bottom at \(10\ \text{m s}^{-1}\). Find the vertical height it descended. | Whole class, then pairs |
| 13 to 21 min | Teach and check: The work-energy principle | When friction, resistance or a driving force does work, keep an energy account. Model the worked example Down a rough slope on the board, one line at a time. | Pairs |
| 21 to 30 min | Teach and check: Driving uphill | With a driving force, its work goes in on the left of the account. Model the worked example A cyclist on a hill on the board, one line at a time. | Connect-Extend-Challenge |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 23 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Driving uphill, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Connect-Extend-Challenge
After the new idea, each student writes three lines: how it connects to what they knew, how it extends their thinking, and what still challenges them. Pairs compare, then the challenges become the plenary.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Adding speeds instead of energies. | The ball thrown at 12 and gaining 10 from the fall does NOT land at 22: energies add, so \(v^2 = 144 + 100\). |
| \(\tfrac{1}{2}m(25 - 15)^2\) is NOT the change in KE. | Square each speed first: \(\tfrac{1}{2}m(25^2 - 15^2)\). |
| Watch the sign of the KE change. | Here the cyclist SLOWS down, so the lost KE (1560 J) helps: \(W = 18\,000 - 1560\), not \(18\,000 + 1560\). |
Kinetic and potential energy, the work-energy principle: worksheet
6 questions, 23 marks. Written in the style of the papers; no past paper question is reproduced.
A ball of mass 0.5 kg moves at \(8\ \text{m s}^{-1}\) at a height of 3 m above the ground. Find its kinetic energy and its potential energy relative to the ground.
A child of mass 30 kg slides from rest down a smooth curved slide, descending a vertical height of 4 m. Find the speed at the bottom.
A stone is thrown from the top of a cliff 20 m above the sea with speed \(15\ \text{m s}^{-1}\). Ignoring air resistance, find the speed with which it hits the sea.
A particle of mass 2 kg is projected at \(10\ \text{m s}^{-1}\) up a line of greatest slope of a rough plane with \(\sin\alpha = 0.6\). The coefficient of friction is 0.25. Use energy to find (a) the distance it travels up the plane, (b) its speed when it slides back to its starting point.
A car of mass 1000 kg freewheels along a level road, its speed falling from \(25\ \text{m s}^{-1}\) to \(15\ \text{m s}^{-1}\) over 200 m. Assuming a constant resistance, find the resistance and how much further the car travels before stopping.
A roller-coaster car of mass 500 kg starts from rest at A, 30 m above the ground. It passes B, 12 m above the ground, then C, 20 m above the ground. Between A and B the resistances do 15 000 J of work against the car; between B and C the track may be taken as smooth with no resistance. Find the speed of the car at B and at C.
Answer key and marking notes
- KE \(= \tfrac{1}{2}(0.5)(64) = 16\) J.
- PE \(= 0.5 \times 10 \times 3 = 15\) J.
- \(30(10)(4) = \tfrac{1}{2}(30)v^2\), so \(v^2 = 80\).
- \(v = 8.94\ \text{m s}^{-1}\).
- \(\tfrac{1}{2}v^2 = \tfrac{1}{2}(15^2) + 10(20)\).
- \(v^2 = 225 + 400 = 625\).
- \(v = 25\ \text{m s}^{-1}\).
- Friction \(= 0.25 \times 20 \times 0.8 = 4\) N.
- (a) Up a distance \(d\): \(\tfrac{1}{2}(2)(100) = 20(0.6)d + 4d\), so \(100 = 16d\) and \(d = 6.25\) m.
- (b) Down: PE lost \(= 12 \times 6.25 = 75\) J; work against friction \(= 4 \times 6.25 = 25\) J.
- \(\tfrac{1}{2}(2)v^2 = 75 - 25 = 50\), so \(v = 7.07\ \text{m s}^{-1}\).
- KE lost \(= \tfrac{1}{2}(1000)(625 - 225) = 200\,000\) J, so \(R = 1000\) N.
- Remaining KE \(= \tfrac{1}{2}(1000)(225) = 112\,500\) J.
- Further distance \(= 112\,500 \div 1000 = 112.5\) m.
- A to B: PE lost \(= 500 \times 10 \times 18 = 90\,000\) J.
- \(\tfrac{1}{2}(500)v_B^2 = 90\,000 - 15\,000 = 75\,000\), so \(v_B^2 = 300\) and \(v_B = 17.3\ \text{m s}^{-1}\).
- B to C: rises 8 m, so \(\tfrac{1}{2}v_C^2 = \tfrac{1}{2}(300) - 10(8)\).
- \(v_C^2 = 140\), so \(v_C = 11.8\ \text{m s}^{-1}\).
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
A child starts from rest at the top of a smooth slide and descends a vertical height of 5 m. Find the child's speed at the bottom.
- A\(10\ \text{m s}^{-1}\)
- B\(7.07\ \text{m s}^{-1}\)
- C\(100\ \text{m s}^{-1}\)
- D\(3.16\ \text{m s}^{-1}\)
A car of mass 1500 kg speeds up from \(10\ \text{m s}^{-1}\) to \(20\ \text{m s}^{-1}\). Find the increase in its kinetic energy.
- A225 000 J
- B75 000 J
- C300 000 J
- D450 000 J
A ball is thrown with speed \(8\ \text{m s}^{-1}\) from a point 15 m above the ground. Ignoring air resistance, find its speed when it hits the ground, whatever its direction of projection.
- A\(19.1\ \text{m s}^{-1}\)
- B\(25.3\ \text{m s}^{-1}\)
- C\(17.3\ \text{m s}^{-1}\)
- D\(14.6\ \text{m s}^{-1}\)
For a particle moving under several forces, when is the total mechanical energy (kinetic plus gravitational potential) conserved?
- AWhen the only force that does work on it is its weight
- BWhenever it moves at constant speed
- CWhenever it moves on a horizontal surface
- DAlways, because energy cannot be created or destroyed
A particle of mass 3 kg slides 6 m down a line of greatest slope of a rough plane inclined at \(\alpha\), where \(\sin\alpha = 0.6\). The coefficient of friction is 0.3. Its speed at the start is \(2\ \text{m s}^{-1}\). Use the work-energy principle to find its speed after the 6 m.
- A\(6.87\ \text{m s}^{-1}\)
- B\(8.72\ \text{m s}^{-1}\)
- C\(6.32\ \text{m s}^{-1}\)
- D\(6.57\ \text{m s}^{-1}\)
A car of mass 900 kg freewheels along a level road, slowing from \(20\ \text{m s}^{-1}\) to \(12\ \text{m s}^{-1}\) over a distance of 160 m. Find the constant resistance.
- A720 N
- B180 N
- C1440 N
- D45 N
A cyclist and bicycle of total mass 75 kg ride 400 m along a road that rises 20 m vertically. The speed falls from \(9\ \text{m s}^{-1}\) to \(4\ \text{m s}^{-1}\) and there is a constant resistance of 30 N. Find the work done by the cyclist.
- A24 600 J
- B29 400 J
- C12 600 J
- D27 000 J
A particle is projected up a line of greatest slope of a rough plane inclined at \(\alpha\), where \(\sin\alpha = 0.28\) and \(\cos\alpha = 0.96\), with speed \(10\ \text{m s}^{-1}\). The coefficient of friction is 0.25. Find its speed when it returns to its starting point, or decide that it stays at rest.
- A\(2.77\ \text{m s}^{-1}\)
- B\(10\ \text{m s}^{-1}\)
- C\(7.34\ \text{m s}^{-1}\)
- DIt stays at rest
Quiz key
\(\tfrac{1}{2}v^2 = 10 \times 5\), so \(v^2 = 100\) and \(v = 10\ \text{m s}^{-1}\). The shape of the slide does not matter because it is smooth. The tempting 7.07 drops the \(\tfrac{1}{2}\) (\(v^2 = gh = 50\)). 100 is \(v^2\), not \(v\), and 3.16 forgets \(g\).
\(\tfrac{1}{2} \times 1500 \times (20^2 - 10^2) = 750 \times 300 = 225\,000\) J. The tempting 75 000 J uses \(\tfrac{1}{2}m(20 - 10)^2\), the KE of the change in speed, which is not the change in KE. 300 000 J is the final KE only, and 450 000 J forgets the \(\tfrac{1}{2}\).
\(v^2 = 8^2 + 2 \times 10 \times 15 = 64 + 300 = 364\), so \(v = 19.1\ \text{m s}^{-1}\). The tempting 25.3 adds the speeds \(8 + 17.3\); energies add, not speeds. 17.3 ignores the initial speed, and 14.6 drops the 2 in \(2gh\).
KE \(+\) PE stays constant when only gravity does work; other forces may act if they do no work (such as a normal reaction on a smooth surface). The tempting "always" confuses total energy, which is conserved, with MECHANICAL energy: friction turns some into heat. At constant speed up a hill, PE increases while KE stays the same, so KE \(+\) PE is not constant; on a rough horizontal surface friction removes KE.
Initial KE \(= 6\) J; PE lost \(= 30 \times 3.6 = 108\) J; work against friction \(= 0.3 \times 24 \times 6 = 43.2\) J. So \(\tfrac{1}{2} \times 3v^2 = 6 + 108 - 43.2 = 70.8\), giving \(v^2 = 47.2\) and \(v = 6.87\ \text{m s}^{-1}\). The tempting 8.72 ignores friction. 6.32 uses friction \(\mu mg\) instead of \(\mu mg\cos\alpha\), and 6.57 forgets the initial KE.
KE lost \(= \tfrac{1}{2} \times 900 \times (20^2 - 12^2) = 450 \times 256 = 115\,200\) J. Then \(R \times 160 = 115\,200\), so \(R = 720\) N. The tempting 180 N uses \(\tfrac{1}{2}m(20 - 12)^2\) for the KE change. 1440 N forgets the \(\tfrac{1}{2}\), and 45 N uses the change in speed without squaring at all.
PE gained \(= 75 \times 10 \times 20 = 15\,000\) J; work against resistance \(= 30 \times 400 = 12\,000\) J; KE lost \(= \tfrac{1}{2} \times 75 \times (81 - 16) = 2437.5\) J. Work by cyclist \(= 15\,000 + 12\,000 - 2437.5 = 24\,562.5 \approx 24\,600\) J. The tempting 29 400 J ADDS the lost KE; the cyclist slowing down means the KE lost helps. 12 600 J leaves out the resistance, and 27 000 J ignores the change in KE.
Up: per unit mass, \(\tfrac{1}{2} \times 100 = d(10 \times 0.28 + 0.25 \times 10 \times 0.96) = 5.2d\), so \(d = 9.615\) m. \(\tan\alpha = 0.292 > 0.25\), so it slides back. Down: \(\tfrac{1}{2}v^2 = d(2.8 - 2.4) = 3.846\), so \(v = 2.77\ \text{m s}^{-1}\). The tempting \(10\ \text{m s}^{-1}\) ignores the work done against friction both ways. 7.34 ignores friction on the way down only, and "stays at rest" would need \(\mu \ge \tan\alpha\).
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.