Power is the rate of doing work. For a vehicle working at a fixed power, the driving force is \(P/v\): huge at low speed, shrinking as the speed rises, until it only just matches the resistance and the vehicle reaches its top speed.
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.
Fig. 1: driving force F = P/v for a 30 kW engine (to scale) against a constant 1200 N resistance. At 15 m s⁻¹ there is an 800 N resultant; at 25 m s⁻¹ the forces balance and the car can go no faster
Section 01
Rate of doing work.
In the lab, set the road slope to zero, start the car and watch the speed-time graph level off. Raise the engine power and the top speed rises; tilt the road uphill and it falls. The lab uses a drag that grows with speed; Paper 4 questions usually give a constant resistance, but the idea of a top speed is the same.
Average power \(= \dfrac{\text{work done}}{\text{time}}\). Watts; 1 kW \(= 1000\) W.
\(P = Fv\): from work \(= Fd\) per second, with \(d\) per second \(= v\). \(F\) is the DRIVING force, not the resultant.
At a given power, the driving force falls as speed rises (Fig. 1).
Worked example Average power
A motor lifts a load of mass 50 kg through a vertical height of 12 m at constant speed in 8 s. Find the power of the motor.
Work done against gravity \(= 50 \times 10 \times 12 = 6000\) J.
Power \(= \dfrac{6000}{8}\).
Answer: 750 W
Worked example Power from force and speed
A cyclist moves at a steady \(6\ \text{m s}^{-1}\) against a total resistance of 50 N. Find the cyclist's power output.
Steady speed, so driving force \(=\) resistance \(= 50\) N.
\(P = Fv = 50 \times 6\).
Answer: 300 W
Examiner trap
Power must be in WATTS before you use \(F = P/v\). An engine of 30 kW gives \(F = 30\,000/v\), not \(30/v\).
Section 02
Top speed and acceleration.
At any instant, find the driving force from \(F = P/v\), then use Newton's second law. At top speed the acceleration is zero, so driving force equals resistance.
Worked example Maximum speed on the level
A car's engine works at 30 kW against a constant resistance of 1200 N on a level road. Find its maximum speed.
At top speed \(a = 0\), so \(\dfrac{30\,000}{v} = 1200\).
\(v = \dfrac{30\,000}{1200}\).
Answer: \(25\ \text{m s}^{-1}\)
Worked example Acceleration at an instant
The car has mass 1200 kg. Find its acceleration at the instant its speed is \(15\ \text{m s}^{-1}\).
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 \(R\).
Top speed: \(\dfrac{P}{40} = R\). At 20: \(\dfrac{P}{20} - R = 500\), so \(\dfrac{P}{20} - \dfrac{P}{40} = 500\), \(\dfrac{P}{40} = 500\). \(P = 20\,000\) W (20 kW) and \(R = 500\) N.
Examiner trap
\(P = Fv\) uses the DRIVING force. Writing \(P = mav\) (the resultant) at 15 m/s gives \(a = 1.67\) instead of 0.667.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
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Which unit is the same as a watt?
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A driving force of 400 N moves a car at \(15\ \text{m s}^{-1}\). Find the power.
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A car travels at its maximum speed on a level road. What is true?
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A machine does 2400 J of work in 8 s. Find its average power.
Section 03
Up and down hills.
On a slope the weight component \(mg\sin\alpha\) joins the resistance going uphill and helps going downhill. Hills are often given as \(\sin\alpha = 0.05\) ("1 in 20").
Worked example Uphill
The 1200 kg car (30 kW, resistance 1200 N) drives up a hill inclined at \(\alpha\), where \(\sin\alpha = 0.05\). Find its maximum speed, and its acceleration when its speed is \(10\ \text{m s}^{-1}\).
Weight component down the hill \(= 12\,000 \times 0.05 = 600\) N.
Top speed: \(\dfrac{30\,000}{v} = 1200 + 600 = 1800\), so \(v = 16.7\ \text{m s}^{-1}\).
At 10: driving force \(= 3000\) N, so \(3000 - 1800 = 1200a\).
A lorry of mass 8000 kg descends a hill with \(\sin\alpha = 0.02\) against a resistance of 4000 N, with its engine working at 120 kW. Find its maximum speed.
Weight component down the hill \(= 80\,000 \times 0.02 = 1600\) N, which helps.
At top speed: \(\dfrac{120\,000}{v} + 1600 = 4000\).
\(\dfrac{120\,000}{v} = 2400\), so \(v = 50\).
Answer: \(50\ \text{m s}^{-1}\)
Examiner trap
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.
Stage 2 · Exam
Exam-style questions
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A 1200 kg car works at 30 kW against a resistance of 1200 N on a level road. Find its acceleration when its speed is \(15\ \text{m s}^{-1}\).
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The same car climbs a hill with \(\sin\alpha = 0.05\). Find its maximum speed.
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A motor lifts a 50 kg load through 12 m at constant speed in 8 s. Find its average power.
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A 1000 kg car at constant power has top speed \(40\ \text{m s}^{-1}\) against a constant resistance, and acceleration \(0.5\ \text{m s}^{-2}\) at \(20\ \text{m s}^{-1}\). Find the power.
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