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Power and efficiency (QCE Physics Unit 2)

Syllabus dot point

“Define power as the rate of doing work or transferring energy, and apply P=W/t=FvP = W / t = Fv to mechanical systems, including efficiency calculations”

8 min read

Quick answer

Power is the rate of doing work or transferring energy, defined as P=W/tP = W/t and equal to FvFv for a constant force, and efficiency is the dimensionless ratio of useful energy output to total energy input (η=Pout/Pin\eta = P_{\text{out}}/P_{\text{in}}), with the difference appearing as friction, heat, sound or resistive losses.

Jump to a section
  1. What this dot point is asking
  2. Power
  3. Constant-power problems
  4. Efficiency
  5. Connection to the rest of Unit 2
  6. Examples in context
  7. Try this
  8. Exam-style questions

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):

P=WtP = \frac{W}{t}

SI unit: watt (W = J s−1^{-1}). For a force pushing an object at constant velocity:

P=FvP = F v

where FF 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:

F=PvF = \frac{P}{v}

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 vv is large.

At top speed (steady velocity), driving force equals total resistive force, so P=Fresistivevmax⁡P = F_{\text{resistive}} v_{\max}.

Efficiency

Efficiency is the dimensionless ratio of useful output to total input:

η=Euseful outEtotal in=PoutPin\eta = \frac{E_{\text{useful out}}}{E_{\text{total in}}} = \frac{P_{\text{out}}}{P_{\text{in}}}

Expressed as a fraction or a percentage. Always less than 11 in real machines; the rest is energy lost as heat, sound, friction, or electrical resistance.

Engine power and efficiency at steady speed

A car of mass 12001200 kg travels at a steady 2525 m s−1^{-1} against a total resistive force of 600600 N. Find the engine power output, and the efficiency if the engine consumes 5050 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:

Fdrive=Fresistive=600 NF_{\text{drive}} = F_{\text{resistive}} = 600 \text{ N}

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 P=FvP = Fv:

Pout=Fv=(600)(25)=15 000 W=15 kWP_{\text{out}} = Fv = (600)(25) = 15\,000 \text{ W} = 15 \text{ kW}

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 5050 kW:

η=PoutPin=1550=0.30=30%\eta = \frac{P_{\text{out}}}{P_{\text{in}}} = \frac{15}{50} = 0.30 = 30\%

This figure is typical for a petrol engine; the remaining 70%70\% is lost as heat in the combustion and cooling systems.

Final answer: engine power output =15= 15 kW; efficiency =30%= 30\%.

Common traps
Treating P=FvP = Fv as instantaneous when it is not
P=FvP = Fv 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−1^{-1}). Joules are energy. A 100100 W bulb running for 11 s consumes 100100 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 (P=W/tP = W/t converts a kinematic work calculation into a rate) and to motion graphs (constant-power motion gives a non-linear vv-tt curve because acceleration depends on vv). 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 P=FvP = Fv by hand.

Examples in context

Example 1. A Bundaberg sugar mill's bagasse-fired boiler delivers 40 MW40 \text{ MW} thermal at 8080 per cent boiler efficiency. The downstream turbine (3535 per cent thermal-to-mechanical) produces 40×0.80×0.35=11.2 MW40 \times 0.80 \times 0.35 = 11.2 \text{ MW} shaft power, of which the generator converts 9595 per cent to electrical: 10.6 MW10.6 \text{ MW} delivered. Overall plant efficiency is 26.626.6 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 200 kW200 \text{ kW} traction power at 14 m s−114 \text{ m s}^{-1}, so it exerts F=P/v=14300 NF = P/v = 14300 \text{ N} on the rail. Electrical input is 230 kW230 \text{ kW} from the 750 V750 \text{ V} DC bus, giving motor efficiency 8787 per cent. Engineers monitor the 1313 per cent loss (mostly I2RI^2 R 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. P=W/tP = W/t; also P=FvP = Fv at constant velocity.

Q2. A motor lifts a 200 kg200 \text{ kg} load 15 m15 \text{ m} in 20 s20 \text{ s}. Calculate the useful output power. If input power is 1.8 kW1.8 \text{ kW}, calculate the efficiency. [3 marks]

  • Cue. W=mgh=29400 JW = mgh = 29400 \text{ J}; Pout=1470 WP_{out} = 1470 \text{ W}; efficiency =0.817=81.7= 0.817 = 81.7 per cent.

Q3. A Cairns tram induction motor delivers 200 kW200 \text{ kW} at 14 m s−114 \text{ m s}^{-1}. (a) Calculate the traction force. (b) Given 230 kW230 \text{ kW} 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) 14300 N14300 \text{ N}; (b) 8787 per cent, 30 kW30 \text{ kW} lost; (c) I2RI^2 R 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 marks
An electric motor lifts a 200200 kg load at a constant speed of 0.500.50 m s−1^{-1}. Using g=9.8g = 9.8 m s−2^{-2}, calculate (a) the mechanical power output. If the motor draws 12001200 W from the supply, find (b) its efficiency.
Show worked answer →

(a) Mechanical power output. At constant speed the lifting force equals weight, F=mg=(200)(9.8)=1960F = mg = (200)(9.8) = 1960 N.

Pout=Fv=(1960)(0.50)=980P_{\text{out}} = F v = (1960)(0.50) = 980 W.

(b) Efficiency.

η=PoutPin=9801200=0.817=81.7%\eta = \frac{P_{\text{out}}}{P_{\text{in}}} = \frac{980}{1200} = 0.817 = 81.7\%.

Markers reward the recognition that at constant speed F=mgF = mg, the substitution into P=FvP = Fv, and a clear ratio statement of efficiency.

Practise this

Sources & how we know this

ExamExplained