Power and efficiency (QCE Physics Unit 2)
“Define power as the rate of doing work or transferring energy, and apply to mechanical systems, including efficiency calculations”
Power is the rate of doing work or transferring energy, defined as and equal to for a constant force, and efficiency is the dimensionless ratio of useful energy output to total energy input (), with the difference appearing as friction, heat, sound or resistive losses.
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What this dot point is asking
QCAA wants you to use power as the rate quantity that links work and time, and to use efficiency as the ratio of useful energy output to total energy input. Both ideas appear in IA1 stimulus and EA Paper 1, often as small calculations attached to a larger motion or energy problem.
Power
Power is the rate at which work is done (or energy is transferred):
SI unit: watt (W = J s). For a force pushing an object at constant velocity:
where is the component of force in the direction of motion. This form is useful when the force is constant and the speed is steady (cruising car, elevator at constant speed, conveyor belt).
Constant-power problems
When a vehicle delivers constant power, the driving force decreases as speed increases:
This is why a car has plenty of acceleration at low speeds but accelerates only slightly near top speed: the engine power is the same but is large.
At top speed (steady velocity), driving force equals total resistive force, so .
Efficiency
Efficiency is the dimensionless ratio of useful output to total input:
Expressed as a fraction or a percentage. Always less than in real machines; the rest is energy lost as heat, sound, friction, or electrical resistance.
A car of mass kg travels at a steady m s against a total resistive force of N. Find the engine power output, and the efficiency if the engine consumes kW of fuel energy.
Step 1: Identify the driving force at constant velocity
At steady speed there is no acceleration, so by Newton's first law the net force is zero. The driving force from the engine must exactly balance the total resistive force:
Step 2: Calculate the mechanical power output
Power is the rate at which the driving force does work. For a constant force and constant speed, use :
Step 3: Calculate the efficiency
Efficiency is the ratio of useful mechanical output to total energy input from fuel. Here the fuel power is given as kW:
This figure is typical for a petrol engine; the remaining is lost as heat in the combustion and cooling systems.
Final answer: engine power output kW; efficiency .
- Treating as instantaneous when it is not
- gives instantaneous power for a constant force; it gives average power for an average force.
- Mixing up watts and joules
- Watts are power (J s). Joules are energy. A W bulb running for s consumes J of electrical energy.
- Computing efficiency from forces, not energies
- Efficiency compares energy or power, not raw forces. The lifting force and pulling force in a pulley system can differ by a factor of two while still delivering the same work.
- Forgetting that efficiency is dimensionless
- Both numerator and denominator must use the same unit (joules with joules, watts with watts).
Connection to the rest of Unit 2
Power links to the work and mechanical energy dot point ( converts a kinematic work calculation into a rate) and to motion graphs (constant-power motion gives a non-linear - curve because acceleration depends on ). In Unit 3, the same power-efficiency framework reappears for transformers and AC transmission.
Try it: For a horizontal-motion power calculation, use the SUVAT calculator to get the steady velocity, then apply by hand.
Examples in context
Example 1. A Bundaberg sugar mill's bagasse-fired boiler delivers thermal at per cent boiler efficiency. The downstream turbine ( per cent thermal-to-mechanical) produces shaft power, of which the generator converts per cent to electrical: delivered. Overall plant efficiency is per cent. The QCAA Unit 2 dot point's efficiency chain is exactly this product of stage efficiencies, a standard EA Paper 1 question.
Example 2. A Cairns light-rail induction motor produces traction power at , so it exerts on the rail. Electrical input is from the DC bus, giving motor efficiency per cent. Engineers monitor the per cent loss (mostly in windings, magnetic core loss, mechanical friction) to schedule maintenance; departures from baseline efficiency are a QCAA-style stimulus for IA1 data analysis.
Try this
Q1. Define power and state two equivalent algebraic forms. [2 marks]
- Cue. ; also at constant velocity.
Q2. A motor lifts a load in . Calculate the useful output power. If input power is , calculate the efficiency. [3 marks]
- Cue. ; ; efficiency per cent.
Q3. A Cairns tram induction motor delivers at . (a) Calculate the traction force. (b) Given electrical input, determine the efficiency and the power lost. (c) Identify two physical loss mechanisms with reasoning. [2+3+3 marks; ISMG: Knowledge and conceptual understanding, Evaluation]
- Cue. (a) ; (b) per cent, lost; (c) winding loss, magnetic core hysteresis-eddy losses.
Exam-style questions
Questions in the style of QCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.
Year 11 SAC-style4 marksAn electric motor lifts a kg load at a constant speed of m s. Using m s, calculate (a) the mechanical power output. If the motor draws W from the supply, find (b) its efficiency.Show worked answer →
(a) Mechanical power output. At constant speed the lifting force equals weight, N.
W.
(b) Efficiency.
.
Markers reward the recognition that at constant speed , the substitution into , and a clear ratio statement of efficiency.