Lesson 3: Work, Energy and Power

Energy is the great bookkeeping device of physics: instead of tracking forces through every instant of a motion, we count energy in and energy out. This lesson covers Theme A.3 — work, kinetic and gravitational potential energy, power and efficiency — with elastic potential energy and the spring force met again in SHM (Lesson 6).

Work and energy

Work is done when a force moves its point of application: W = Fs cos θ, where θ is the angle between the force and the displacement (data booklet). Only the component of force along the displacement does work. The work done by a resultant force equals the change in kinetic energy: W = ΔEk, with Ek = ½mv². Gravitational potential energy changes by ΔEp = mgΔh near Earth's surface. For a spring, the force is F = kx (Hooke's law) and the stored elastic potential energy is the area under the force–extension graph, E = ½kx².

Conservation of energy: in an isolated system the total energy is constant; when friction or drag acts, mechanical energy is transferred to internal (thermal) energy, and the "lost" amount equals the work done against friction.

Power and efficiency

Power is the rate of energy transfer, P = ΔE/Δt = Fv (for a force driving motion at speed v). Efficiency = useful output energy (or power) ÷ total input. Efficiencies in Paper 2 often combine several steps: an electric motor driving a pump has electrical input, motor losses, and useful gravitational potential gain — read carefully which stages the question includes.

Worked problem 1 — rollercoaster with friction

A 58 kg rider in a cart is released from rest at the top of a 32 m drop. At the bottom the speed is measured as 23 m s⁻¹ (not the ideal value). Calculate (a) the ideal speed with no friction and (b) the energy transferred to thermal energy by friction and drag.

(a) mgh = ½mv²  →  v = √(2gh) = √(2 × 9.81 × 32) = √(627.8) = 25 m s⁻¹
(b) Initial Ep = mgh = 58 × 9.81 × 32 = 18 207 J ≈ 18.2 kJ
    Final Ek = ½ × 58 × 23² = 15 341 J ≈ 15.3 kJ
    Energy to thermal = 18.2 − 15.3 = 2.9 kJ

The mass is needed here because the question asks for an energy, not just a speed — a favourite examiner variation.

Worked problem 2 — pump power and efficiency

A pump raises 250 kg of water through 18 m in 60 s. Its electrical input power is 1.1 kW. Calculate the useful power and the efficiency.

Useful energy per minute = mgh = 250 × 9.81 × 18 = 44 145 J
Useful power = 44 145 / 60 = 736 W ≈ 0.74 kW
Efficiency = 736 / 1100 = 0.67 → 67%

Practical worked example — your own power output

Running up stairs is the classic measurement. A 62 kg student climbs a vertical height of 3.2 m in 4.1 ± 0.2 s:

P = mgh / t = 62 × 9.81 × 3.2 / 4.1 = 1946 / 4.1 = 475 W

Human reaction time on the stopwatch contributes ±0.2 s, about 5%, dwarfing the uncertainties in mass and height — so the honest result is 475 ± 25 W. Repeating the climb and timing 5 runs reduces the random component. Values of 400–800 W for a few seconds are typical; sustained human power is only ~100 W, a contrast worth discussing in evaluation.

Nature of Science

Energy is not a substance but an accounting system — yet it is one of the most powerful ideas in science because the books always balance. When beta decay seemed to violate energy conservation, Pauli proposed the neutrino rather than abandon the law; the particle was detected 25 years later. Conservation laws guide discovery.

Real-world application

Regenerative braking in electric vehicles converts kinetic energy back into stored chemical energy instead of wasting it as heat in brake pads; typical recovery is 60–70% of the braking energy, which is why EVs are especially efficient in city driving with repeated stops.

Common mistakes (examiner style)

• Using Ep = mgh with h measured from the ground when the motion starts above it — only the change in height matters.
• Calculating work done by friction as μN × time instead of μN × distance.
• Forgetting that work done by a force at right angles to the motion is zero (for example, the normal reaction, or centripetal force).
• Giving efficiency greater than 1 by inverting output and input.
• In P = Fv, using the top speed with the initial force — F and v must be simultaneous values.

IB Exam Experience

Paper 1-style questions. 1. A constant force of 50 N pushes a box 8.0 m along the direction of the force. The work done is: A. 6.3 J B. 50 J C. 400 J D. 3200 J. Key: C — W = Fs = 50 × 8.0 = 400 J. 2. A car of mass 900 kg doubles its speed from 12 m s⁻¹ to 24 m s⁻¹. Its kinetic energy increases by a factor of: A. 2 B. 4 C. 8 D. 16. Key: B — Ek ∝ v². 3. A 2.0 kW motor lifts a load at constant speed 0.50 m s⁻¹. The weight of the load is: A. 1000 N B. 2000 N C. 4000 N D. 8000 N. Key: C — P = Fv → F = 2000/0.50 = 4000 N (weight = force at constant speed).

Paper 2-style structured question (7 marks). A cyclist produces a useful power of 240 W while riding at 9.0 m s⁻¹ on a level road. (a) Calculate the total resistive force acting on the cyclist. [2] (b) The cyclist stops pedalling and slows under the same resistive force (mass of cyclist and bicycle 78 kg). Calculate the initial deceleration. [2] (c) On a hill the cyclist rides at 4.0 m s⁻¹ up a slope of 4.0°, still producing 240 W. Explain, with a calculation, whether all the power now goes against gravity. [3] *Marking notes:* (a) F = P/v = 240/9.0 = 27 N. (b) a = F/m = 27/78 = 0.34 m s⁻². (c) Rate of Ep gain = mgv sin θ = 78 × 9.81 × 4.0 × sin 4.0° = 214 W < 240 W, so no — about 26 W still works against resistance; accept 210–220 W for the Ep rate given rounding. Watch for candidates who use the slope angle with cos or who forget that "at constant speed" means zero resultant force in (a)'s reasoning.

IA idea: Investigate the efficiency of a small electric motor lifting masses: how does efficiency vary with load, and where does the "missing" energy go?

The work–energy theorem as a derivation

The link between work and kinetic energy is not a coincidence: for a constant resultant force, W = Fs = ma × (v² − u²)/(2a) = ½mv² − ½mu², using v² = u² + 2as. For variable forces the same result follows from W = ∫F dx. The practical consequence: energy methods give speeds directly from distances (through heights, extensions and displacements) without ever finding times or accelerations. Choose the tool by what the question gives: distances and heights → energy; times and impulses → momentum; forces and accelerations → Newton.

Springs: energy stored and released

A spring stretched by x stores E = ½kx² (the area under the F–x line). A child of mass 28 kg on a pogo stick compresses its spring (k = 4.2 × 10³ N m⁻¹) by 0.16 m at the bottom of a bounce; the stored energy is ½ × 4.2 × 10³ × 0.16² = 54 J. If all of it returns as gravitational potential energy, the child rises by h = 54/(28 × 9.81) = 0.20 m above the compression point. Real springs return 80–90%: measuring the shortfall is a neat investigation of hysteresis — the loading and unloading F–x graphs enclose a loop whose area is the energy dissipated per cycle.

Efficiency chains in power generation

Real systems chain efficiencies multiplicatively. A coal power station: boiler 0.88 × turbine 0.45 × generator 0.97 × transmission 0.93 = 0.36 overall — the familiar ~36% figure. The lesson for exam questions: when a problem stacks devices, compute stage-by-stage and multiply the fractions; never average them. The same bookkeeping applies to an electric car charged from that station: 0.36 (generation) × 0.90 (charging) × 0.85 (motor) ≈ 0.28 from coal to road — a defensible, data-driven contribution to any energy-policy discussion.

One more energy chain — bouncing ball analysis

A ball dropped from 1.80 m rebounds to 1.25 m, then 0.87 m. Each bounce returns a fraction (1.25/1.80) = 0.69 of the height — and since height ∝ energy, each bounce keeps 69% of the mechanical energy and dissipates 31%. After n bounces the height is 1.80 × 0.69ⁿ: below 0.10 m after 7 bounces. This constant-fraction pattern is the signature of a fixed coefficient of restitution behaviour and appears in crash-test and sports-engineering data alike; plotting ln(height) against bounce number linearises it (Lesson 18), giving the energy-retention fraction from the gradient — a complete mini-investigation hiding in a playground observation.

Energy diagrams: reading the whole story at a glance

A Sankey diagram draws energy flows to scale: arrow widths proportional to energy per second (or per event). For the pump of Worked Problem 2, the input arrow of 1100 W splits into 736 W of useful potential-energy gain and 364 W of losses (friction in the pump, turbulence in the water, sound). Training your eye to sketch a Sankey for any multi-stage device — power station, car journey, heating system — turns efficiency questions into area comparisons, and errors (outputs summing to more than input) become visually impossible to miss. Examiners who ask you to "complete the Sankey" are testing conservation plus proportion in a single picture.