The shape of the lesson
By the end of the lesson, learners can
- Rate of doing work: In the lab, set the road slope to zero, start the car and watch the speed-time graph level off.
- Top speed and acceleration: At any instant, find the driving force from \(F = P/v\), then use Newton's second law.
- Up and down hills: On a slope the weight component \(mg\sin\alpha\) joins the resistance going uphill and helps going downhill.
- Units: power in watts (kW \(\times 1000\)).
- Driving force: \(F = \dfrac{P}{v}\) at the speed in question.
- Hill: \(mg\sin\alpha\) opposes motion uphill and helps downhill.
- Newton's second law: driving force \(-\) resistances \(= ma\).
- Top speed: set \(a = 0\).
One picture every idea returns to
Power \(= \dfrac{\text{work done}}{\text{time taken}}\), in watts (\(1\ \text{W} = 1\ \text{J s}^{-1}\)). For a force in the direction of motion, \(P = Fv\), so the driving force is \(F = \dfrac{P}{v}\). At maximum speed the acceleration is zero and the driving force equals the total resistance.
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, Power is force times speed.. Power is the rate of doing work. 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: Rate of doing work | In the lab, set the road slope to zero, start the car and watch the speed-time graph level off. Project tab 5 of the lab (Work, energy and power lab, power, drag and top speed) and let a student drive it while the class predicts. Model the worked example Average power on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Top speed and acceleration | At any instant, find the driving force from \(F = P/v\), then use Newton's second law. Model the worked example Maximum speed on the level on the board, one line at a time. Your turn: A car of mass 1000 kg works at constant power \(P\) W against a constant resistance \(R\) N. Its top speed on a level road is \(40\ \text{m s}^{-1}\), and at \(20\ \text{m s}^{-1}\) its acceleration is \(0.5\ \text{m s}^{-2}\). Find \(P\) and \(R\). | Pairs |
| 21 to 30 min | Teach and check: Up and down hills | On a slope the weight component \(mg\sin\alpha\) joins the resistance going uphill and helps going downhill. Model the worked example Uphill on the board, one line at a time. | Anticipation / Reaction Guide |
| 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, 25 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 Up and down hills, 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.
Anticipation / Reaction Guide
Before the lesson, students mark five statements true or false. After the teaching they mark them again and justify any change of mind in one sentence each. The statements come from the examiner traps below.
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 |
|---|---|
| Power must be in WATTS before you use \(F = P/v\). | An engine of 30 kW gives \(F = 30\,000/v\), not \(30/v\). |
| \(P = Fv\) uses the DRIVING force. | Writing \(P = mav\) (the resultant) at 15 m/s gives \(a = 1.67\) instead of 0.667. |
| Downhill the weight component is on the SAME side as the driving force. | Adding it to the resistance gives a top speed of 21.4 m/s for the lorry instead of 50 m/s. |
Power and driving force: worksheet
6 questions, 25 marks. Written in the style of the papers; no past paper question is reproduced.
Define power. A car's engine exerts a driving force of 400 N while the car moves at \(15\ \text{m s}^{-1}\). Find the power.
A motor raises a load of mass 50 kg vertically through 12 m at constant speed in 8 s. Find the work done and the power of the motor.
A car of mass 1200 kg has an engine working at 30 kW. It moves on a level road against a constant resistance of 1200 N. Find the maximum speed and the acceleration when the speed is \(15\ \text{m s}^{-1}\).
A car of mass 1000 kg moves on a level road with its engine working at a constant power \(P\) W against a constant resistance \(R\) N. Its maximum speed is \(40\ \text{m s}^{-1}\). When its speed is \(20\ \text{m s}^{-1}\) its acceleration is \(0.5\ \text{m s}^{-2}\). Find \(P\) and \(R\).
A car of mass 1200 kg drives up a straight hill inclined at \(\alpha\) to the horizontal, where \(\sin\alpha = 0.05\). The engine works at 30 kW and the resistance is 1200 N. Find the maximum speed up the hill and the acceleration at the instant the speed is \(10\ \text{m s}^{-1}\).
A lorry of mass 8000 kg travels down a straight hill inclined at \(\alpha\), where \(\sin\alpha = 0.02\). The resistance to motion is a constant 4000 N and the engine works at 120 kW.
(a) Find the maximum speed of the lorry down the hill.
(b) Find its acceleration at the instant its speed is \(30\ \text{m s}^{-1}\).
Answer key and marking notes
- Power is the rate at which work is done.
- \(P = Fv = 400 \times 15 = 6000\) W \(= 6\) kW.
- Work \(= 50 \times 10 \times 12 = 6000\) J.
- Power \(= 6000 \div 8 = 750\) W.
- Top speed: \(30\,000 \div v = 1200\), so \(v = 25\ \text{m s}^{-1}\).
- At 15: \(F = 2000\) N, \(2000 - 1200 = 1200a\).
- \(a = 0.667\ \text{m s}^{-2}\).
- Top speed: \(\dfrac{P}{40} = R\).
- At 20: \(\dfrac{P}{20} - R = 1000 \times 0.5 = 500\).
- Substitute: \(\dfrac{P}{20} - \dfrac{P}{40} = \dfrac{P}{40} = 500\).
- \(P = 20\,000\) (20 kW) and \(R = 500\).
- Weight component \(= 1200 \times 10 \times 0.05 = 600\) N down the hill.
- Top speed: \(\dfrac{30\,000}{v} = 1800\), so \(v = 16.7\ \text{m s}^{-1}\).
- At 10: \(3000 - 1800 = 1200a\), so \(a = 1\ \text{m s}^{-2}\).
- Weight component down the hill \(= 80\,000 \times 0.02 = 1600\) N.
- (a) \(\dfrac{120\,000}{v} + 1600 = 4000\), so \(\dfrac{120\,000}{v} = 2400\) and \(v = 50\ \text{m s}^{-1}\).
- (b) At 30: driving force \(= 4000\) N.
- \(4000 + 1600 - 4000 = 8000a\), so \(a = 0.2\ \text{m s}^{-2}\).
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 cyclist moves at a steady \(8\ \text{m s}^{-1}\) against a total resistance of 40 N. Find the cyclist's power output.
- A320 W
- B5 W
- C160 W
- D2560 W
A motor lifts a load of mass 80 kg vertically through 15 m at constant speed in 12 s. Find the power of the motor.
- A1000 W
- B100 W
- C12 000 W
- D800 W
A car's engine works at a constant 24 kW against a constant resistance of 800 N on a level road. Find the car's maximum speed.
- A\(30\ \text{m s}^{-1}\)
- B\(0.03\ \text{m s}^{-1}\)
- C\(33.3\ \text{m s}^{-1}\)
- D\(0.0333\ \text{m s}^{-1}\)
A car moves along a straight road at its maximum speed with its engine working at constant power. Which statement is true?
- AThe driving force equals the total resistance
- BThe driving force is zero
- CThe power output is zero
- DThe resistance is zero
A car of mass 1000 kg has an engine working at 20 kW against a constant resistance of 600 N on a level road. Find its acceleration at the instant its speed is \(16\ \text{m s}^{-1}\).
- A\(0.65\ \text{m s}^{-2}\)
- B\(1.25\ \text{m s}^{-2}\)
- C\(1.85\ \text{m s}^{-2}\)
- D\(0.6\ \text{m s}^{-2}\)
A car of mass 1200 kg moves on a level road against a constant resistance of \(R\) N, with its engine working at a constant \(P\) W. Its maximum speed is \(40\ \text{m s}^{-1}\), and its acceleration is \(0.5\ \text{m s}^{-2}\) when its speed is \(20\ \text{m s}^{-1}\). Find \(P\) and \(R\).
- A\(P = 24\,000\), \(R = 600\)
- B\(P = 12\,000\), \(R = 300\)
- C\(P = 8000\), \(R = 200\)
- D\(P = 24\,000\), \(R = 1200\)
A car of mass 1200 kg drives up a straight hill inclined at \(\alpha\) to the horizontal, where \(\sin\alpha = 0.05\). Its engine works at 30 kW and the resistance to motion is a constant 900 N. Find the car's maximum speed up the hill.
- A\(20\ \text{m s}^{-1}\)
- B\(100\ \text{m s}^{-1}\)
- C\(33.3\ \text{m s}^{-1}\)
- D\(31.3\ \text{m s}^{-1}\)
A lorry of mass 10 000 kg descends a straight hill inclined at \(\alpha\) to the horizontal, where \(\sin\alpha = 0.03\), against a constant resistance of 5000 N. Its engine works at 100 kW. Find its maximum speed.
- A\(50\ \text{m s}^{-1}\)
- B\(12.5\ \text{m s}^{-1}\)
- C\(20\ \text{m s}^{-1}\)
- D\(33.3\ \text{m s}^{-1}\)
Quiz key
Driving force \(= 40\) N, so \(P = 40 \times 8 = 320\) W. The tempting 5 W divides the force by the speed instead of multiplying. 160 W brings in a \(\tfrac{1}{2}\) from the kinetic energy formula, and 2560 W multiplies by \(v^2\).
Work \(= 80 \times 10 \times 15 = 12\,000\) J, so power \(= 12\,000 \div 12 = 1000\) W (or \(P = Fv = 800 \times 1.25\)). The tempting 12 000 W is the work done, not divided by the time. 100 W forgets \(g\), and 800 is the weight in newtons, a force.
At top speed \(\dfrac{24\,000}{v} = 800\), so \(v = 30\ \text{m s}^{-1}\). The tempting 0.03 leaves the power in kilowatts (\(24 \div 800\)). 33.3 and 0.0333 divide the resistance by the power (\(800 \div 24\) and \(800 \div 24\,000\)), turning \(P = Fv\) upside down.
At maximum speed the acceleration is zero, so the resultant force is zero: driving force \(=\) resistance (plus any weight component on a hill). The tempting "driving force is zero" confuses zero acceleration with zero force; the engine is still working at power \(P = Fv\), with \(F\) balancing the resistance.
Driving force \(= \dfrac{20\,000}{16} = 1250\) N. Resultant \(= 1250 - 600 = 650\) N, so \(a = 0.65\ \text{m s}^{-2}\). The tempting 1.25 uses \(P = mav\), treating the driving force as the resultant and ignoring the resistance. 1.85 adds the resistance, and 0.6 divides the resistance by the mass.
Substituting \(R = \dfrac{P}{40}\): \(\dfrac{P}{20} - \dfrac{P}{40} = 600\), so \(\dfrac{P}{40} = 600\), \(P = 24\,000\) and \(R = 600\). The tempting \(P = 12\,000\) sets \(\dfrac{P}{20} = 600\), forgetting the resistance at 20 m/s. \(P = 8000\) adds the resistance instead of subtracting, and \(R = 1200\) takes the driving force at 20 m/s as the resistance.
Weight component \(= 1200 \times 10 \times 0.05 = 600\) N down the hill. At top speed, \(\dfrac{30\,000}{v} = 900 + 600\), so \(v = 20\ \text{m s}^{-1}\). The tempting 100 subtracts the weight component, which is the downhill case. 33.3 ignores the hill, and 31.3 uses \(m\sin\alpha = 60\) N without \(g\).
Weight component \(= 10\,000 \times 10 \times 0.03 = 3000\) N down the hill, helping. At top speed, driving force \(+ 3000 = 5000\), so the driving force is 2000 N and \(v = \dfrac{100\,000}{2000} = 50\ \text{m s}^{-1}\). The tempting 12.5 adds the weight component to the resistance (8000 N), which is the uphill case. 20 ignores the hill, and 33.3 uses the weight component alone as the driving force.
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.