Work, energy and power is the topic that quietly threads through every other section of the IB Physics syllabus. Energy conservation is your most powerful problem-solving tool — it lets you skip force diagrams, skip kinematics, and jump straight from initial to final state for almost any mechanics question, from a roller coaster to a rolling sphere to a pendulum. Topic A.3 packs in the work-energy theorem, kinetic and potential energy, conservation of mechanical energy with and without friction, power (including $P = Fv$), Sankey-diagram efficiency, and energy density of fuels.
This cheatsheet condenses every formula, trick and trap from Topic A.3 Work, Energy and Power into one page. It covers the work formula and angle traps, the work-energy theorem, kinetic energy (including $E_k = p^2 / 2m$), gravitational and elastic PE, conservation with non-conservative forces, power, Sankey diagrams and efficiency, and a fuel-comparison table for energy density. Scroll to the bottom for the printable PDF and the gated full library.
Cross-reference with the official IB Physics subject page for the full syllabus.
§1 — Work Done A.3
Core formula: $W = Fs\cos\theta$. Units: joule (J). $F$ in newtons, $s$ in metres, $\theta$ is the angle between the force and the displacement.
| $\theta$ | $\cos\theta$ | Work |
|---|---|---|
| 0° | 1 | $W = Fs$ (maximum, force along motion) |
| 90° | 0 | $W = 0$ (force perpendicular) |
| 180° | $-1$ | $W = -Fs$ (force opposes motion) |
Work-energy theorem
$W_{\text{net}} = \Delta E_K$ — the net work done on an object equals its change in kinetic energy.
§2 — Kinetic Energy A.3
$$E_K = \tfrac{1}{2} m v^2 = \dfrac{p^2}{2m}$$
§3 — Potential Energies A.3
Gravitational PE (near Earth's surface)
$\Delta E_p = mg\Delta h$. $\Delta h > 0$ ⇒ PE increases (object moves up).
Elastic PE (Hooke's law spring)
$E_H = \tfrac{1}{2} k (\Delta x)^2$. Always $\geq 0$ (extension or compression both store energy).
§4 — Conservation of Mechanical Energy A.3
$$E_{\text{mech}} = E_K + E_p + E_H$$
Conserved when only conservative forces act (no friction, no air resistance).
With friction or drag: $\Delta E_{\text{mech}} = W_{\text{non-cons}} < 0$ — mechanical energy is lost to thermal energy.
§5 — Power A.3
Units: watt (W) $=$ J s⁻¹.
§6 — Efficiency A.3
$$\eta = \dfrac{E_{\text{output}}}{E_{\text{input}}} = \dfrac{P_{\text{output}}}{P_{\text{input}}} \leq 1$$
§7 — Energy Density A.3
$$\text{Energy density} = \dfrac{E_{\text{released}}}{m_{\text{fuel}}} \quad [\text{J kg}^{-1}]$$
| Fuel | Approx. energy density (MJ kg⁻¹) |
|---|---|
| Hydrogen | ~142 |
| Petrol / gasoline | ~46 |
| Coal | ~27 |
| Lithium-ion battery | ~0.6 |
§8 — Master Formula Summary & Exam Attack Plan A.3 — All sections
- $W = F s \cos\theta$
- $W_{\text{net}} = \Delta E_K$
- $E_K = \tfrac{1}{2} m v^2 = p^2 / (2m)$
- $\Delta E_p = mg\Delta h$ (near Earth)
- $E_H = \tfrac{1}{2} k (\Delta x)^2$
- $E_{\text{mech}} = E_K + E_p + E_H$
- $P = \Delta W / \Delta t = Fv$
- $\eta = E_{\text{out}} / E_{\text{in}}$
| Question trigger | Reach for |
|---|---|
| "Find work done by force $F$" | $W = Fs\cos\theta$ — angle between $F$ and $s$ |
| "Find speed at the bottom of a slope" | $mgh = \tfrac{1}{2} m v^2$ (energy conservation) |
| "Friction acts on the block" | $mgh = \tfrac{1}{2} m v^2 + f \cdot d$ (friction takes energy) |
| "Spring released, find max compression" | $\tfrac{1}{2} k x^2 = \tfrac{1}{2} m v^2$ or $mgh$ |
| "Power of a motor / engine" | $P = Fv$; at terminal velocity $F = $ drag force |
| "Efficiency of a heater / motor" | $\eta = P_{\text{out}} / P_{\text{in}}$; waste $= P_{\text{in}}(1 - \eta)$ |
| "Compare KE of two fragments same $p$" | $E_K = p^2 / (2m)$ — lighter has more KE |
| "How much fuel?" | mass $=$ total energy needed $\div$ energy density |
Worked Example — IB-Style Energy Conservation with Friction
Question (HL Paper 2 style — 7 marks)
A 0.50 kg block is released from rest at the top of a smooth ramp of height 1.20 m. At the bottom of the ramp, it slides onto a horizontal rough surface (coefficient of dynamic friction $\mu_d = 0.35$) and eventually comes to rest. Take $g = 9.81$ m s⁻². Calculate (a) the speed of the block at the bottom of the ramp, and (b) the distance it slides on the rough surface before stopping.
Solution
- Conservation of energy on the smooth ramp: $mgh = \tfrac{1}{2} m v^2$. (M1)
- Cancel $m$: $v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 1.20}$. (M1)
- $v = 4.85$ m s⁻¹. (A1) [part (a)]
- On the rough surface, friction force $f = \mu_d m g = 0.35 \times 0.50 \times 9.81 = 1.717$ N. (B1)
- Work-energy theorem: $f \cdot d = \tfrac{1}{2} m v^2$ ⇒ $d = \dfrac{\tfrac{1}{2} m v^2}{f}$. (M1)
- $d = \dfrac{0.5 \times 0.50 \times 4.85^2}{1.717} = \dfrac{5.88}{1.717} = 3.43$ m. (M1)(A1) [part (b)]
Examiner's note: A faster route is to combine both stages in one energy equation: $mgh = f \cdot d$, so $d = \dfrac{gh}{\mu_d g} = \dfrac{h}{\mu_d} = \dfrac{1.20}{0.35} = 3.43$ m. Notice the mass cancels. Always check: if mass cancels in your final answer, you've probably done it right.
Common Student Questions
Why does the normal force do no work on a horizontal surface?
What angle do I use in $W = Fs\cos\theta$?
What happens to KE if I double the speed?
When can I use $\Delta E_p = mg\Delta h$?
Can a machine ever have efficiency greater than 1?
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