IB Physics HL A.3 Work, Energy & Power — Complete Cheatsheet
Work, kinetic and potential energy, conservation of mechanical energy, power, efficiency and energy density — every formula, trick and trap from IB Physics HL Topic A.3. Hand-built by an IBO-certified Singapore tutor with 15+ years of IB experience.
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.
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)
TrickOnly the component of force along the displacement does work. The normal force and weight on a horizontal surface do zero work; centripetal force does zero work in circular motion (always perpendicular to $v$).
Trap$\theta$ is the angle between $\vec{F}$ and $\vec{s}$, NOT the angle of the incline or the angle a rope makes with the floor. Always sketch both vectors at a common origin and measure between them.
Work-energy theorem
$W_{\text{net}} = \Delta E_K$ — the net work done on an object equals its change in kinetic energy.
Only the force component along the motion does work. With the force at angle $\theta$ to the displacement, $W = (F\cos\theta)\,s$.
From the Photon question bank
What examiners actually test on this concept:
W = Fs cos θAngle between F and sZero-work forcesWork–energy theoremArea under F–s
The angle trap is the biggest lost mark in the topic.very frequent $\theta$ is between $\vec F$ and $\vec s$ — never the incline angle. A load dragged horizontally by a rope at $30^\circ$ above the floor gets $W = Fs\cos30^\circ$.
Zero-work forces earn easy marks: the normal force, weight on the flat, and the centripetal force all do no work because they act perpendicular to the motion.
Work = area under a force–distance graph even when $F$ varies — split into triangles and rectangles, or count squares.
§2 — Kinetic Energy A.3
$$E_K = \tfrac{1}{2} m v^2 = \dfrac{p^2}{2m}$$
Note$E_K = p^2 / (2m)$ is essential for comparing energies of two objects that have the same momentum (e.g. fragments of an explosion).
TrickSame momentum ⇒ the lighter mass has more KE. Same KE ⇒ the lighter mass has less momentum.
TrapDoubling the speed quadruples KE. Doubling the mass only doubles KE. This is why braking distance scales with $v^2$.
From the Photon question bank
What examiners actually test on this concept:
½mv²p²/2m formv² scalingWork–energy theorem
Doubling the speed quadruples KE.frequent This is why braking distance scales with $v^2$ — a recurring conceptual MCQ.
Use $E_K = p^2/2m$ when two bodies share momentum (explosion fragments): the lighter one carries more KE.
$W_{\text{net}} = \Delta E_K$ short-cuts many problems — get the speed change straight from the net work, no SUVAT.
§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).
Trap$\Delta E_p = mg\Delta h$ is valid only near Earth's surface, where $g$ is approximately constant. For satellites or interplanetary problems use the gravitational-fields formula $E_p = -GMm/r$.
Note$E_H$ equals the area under the $F$–$x$ graph (a triangle): $\tfrac{1}{2} \times k\Delta x \times \Delta x$.
Elastic PE is the area under the $F$–$x$ line. Since $F = k\Delta x$, the triangular area is $E_H = \tfrac12 k\Delta x^2$; the gradient is the spring constant $k$.
From the Photon question bank
What examiners actually test on this concept:
mgΔh½k(Δx)²Elastic PE = areaNear-Earth limit
Elastic PE is the area under $F = k\Delta x$ (a triangle) $= \tfrac12 k\Delta x^2$; extension and compression both store positive energy.
$\Delta E_p = mg\Delta h$ is a near-surface approximation only.frequent For satellites/orbits switch to $E_p = -GMm/r$ (Topic D.1).
$\Delta h$ is the vertical height gained, not the distance travelled along a slope — resolve first.
§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.
TrickAt maximum height, all KE has converted to PE. Use $\tfrac{1}{2}m v_0^2 = mgh_{\max}$ to find $h_{\max}$ in one line — no SUVAT needed.
TrapDo not cancel mass $m$ from energy equations if the problem involves two different masses (e.g. block and bullet, two trolleys).
NoteThe IB Data Booklet phrases this as: "change in total mechanical energy equals work done by non-conservative forces". Friction always reduces mechanical energy.
Energy is conserved as it converts. Rolling down, gravitational PE (gold) turns into KE (navy) while the total mechanical energy stays constant (dashed line).
From the Photon question bank
What examiners actually test on this concept:
KE ↔ PE conversionEnergy bar chartsFriction loss termTwo-mass problems
The workhorse line: $\tfrac12 m v_0^2 = mgh$.very frequent Solve for $h_{\max}$ or the speed at the bottom in one step — no SUVAT needed.
With friction, add the loss: $mgh = \tfrac12 m v^2 + f\,d$. The "missing" mechanical energy has become thermal.
Do not cancel $m$ when two different masses are involved (block + embedded bullet, two trolleys).
§5 — Power A.3
Average power:$P = \dfrac{\Delta W}{\Delta t}$
Instantaneous (along $v$):$P = F v$
General:$P = F v \cos\theta$
Units: watt (W) $=$ J s⁻¹.
TrickAt terminal velocity, driving force $=$ resistive force. Then $P = F_{\text{drive}} \times v_{\text{terminal}}$ gives the minimum engine power required to maintain that speed.
TrickFor a motor lifting a mass at constant speed: $P_{\text{useful}} = mgv$ (rate at which gravitational PE is gained).
Trap$P = Fv$ only when $F$ is parallel to $v$. In general $P = Fv\cos\theta$. Sloping ropes and tow trucks need the cosine.
Power at steady speed. The driving force balances the drag, so the engine delivers $P = Fv$ just to hold the speed.
From the Photon question bank
What examiners actually test on this concept:
P = ΔW/ΔtP = FvTerminal-velocity powerP = Fv cos θ
$P = Fv$ holds only when $F \parallel v$; otherwise $P = Fv\cos\theta$. frequent Sloping tow-ropes need the cosine.
At steady/terminal speed the drive force equals the resistive force, so $P = F_{\text{drag}}\,v$ gives the engine power to hold that speed.
Lifting at constant speed: $P_{\text{useful}} = mgv$, the rate at which gravitational PE is gained.
NoteSankey diagram: arrow width is proportional to energy. The useful output arrow is always narrower than the input arrow; the rest is wasted (almost always thermal).
TrapEfficiency cannot exceed 1 (100%). If you calculate $\eta > 1$, recheck which value you've put in the numerator vs denominator.
Trick$E_{\text{waste}} = E_{\text{input}} (1 - \eta)$ — useful for finding the thermal energy produced by an engine or a motor.
Sankey diagram. Arrow width is proportional to energy; the useful output ($\eta$) is always narrower than the input, and the rest is wasted as thermal energy.
From the Photon question bank
What examiners actually test on this concept:
η = out/in ≤ 1Sankey arrow widthsWaste = in(1−η)Power vs energy form
Efficiency can never exceed 1.frequent If you get $\eta > 1$ you have swapped input and output — recheck.
On a Sankey diagram arrow width $\propto$ energy; the useful arrow is always narrower and the rest (usually thermal) branches off.
Wasted energy $= E_{\text{in}}(1-\eta)$ — the quick route to "how much heat is produced".
TrickMass of fuel needed $=$ total energy required $\div$ energy density. Units: J $\div$ J kg⁻¹ $=$ kg.
NoteHigh energy density means less fuel mass is needed. Batteries have very low energy density compared to fossil fuels — this is why electric cars need large, heavy battery packs.
From the Photon question bank
What examiners actually test on this concept:
Energy density E/mFuel mass = E ÷ densityCompare fuels/batteriesUnit conversion
Fuel mass needed = total energy $\div$ energy density ($\text{J} \div \text{J kg}^{-1} = \text{kg}$) — a clean one-line calculation.
Hydrogen $\gg$ petrol $\gg$ coal $\gg$ Li-ion battery. A battery's low energy density is exactly why EVs carry heavy packs.
Watch the units: data is quoted in MJ kg⁻¹ — convert to J kg⁻¹ before dividing.
§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 $mgh$
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)
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?
Because the displacement is horizontal but the normal force is vertical, so the angle between $F$ and $s$ is 90°. Since $W = Fs\cos\theta$ and $\cos 90° = 0$, the work is zero. The same logic explains why centripetal force does no work in circular motion — it is always perpendicular to $v$, so the displacement and force are perpendicular at every instant.
What angle do I use in $W = Fs\cos\theta$?
The angle between the force vector and the displacement vector — NOT the angle of an incline or any surface. On an inclined plane, if a horizontal force pushes a block up the slope, $\theta$ is the angle between the horizontal $F$ and the slope direction (the displacement), not the slope angle itself. Always draw a diagram with both vectors at the same starting point and measure between them.
What happens to KE if I double the speed?
It quadruples. $E_K = \tfrac{1}{2} m v^2$, so doubling $v$ multiplies KE by $2^2 = 4$. Doubling the mass only doubles KE. This is a Paper 1 MCQ favourite. The same scaling logic comes back in stopping-distance problems: doubling the speed of a car needs four times the braking distance, not twice.
When can I use $\Delta E_p = mg\Delta h$?
Only near Earth's surface, where $g \approx 9.81$ m s⁻² is approximately constant. For satellites, planets, or anything far from the surface, you must use the gravitational-fields formula $E_p = -GMm/r$ (Topic D.1). The standard $mgh$ approximation breaks down once the height is comparable to Earth's radius. For all "roller-coaster" and "ball-rolling-down-slope" problems on Earth, $mgh$ is fine.
Can a machine ever have efficiency greater than 1?
No. Efficiency $\eta = $ useful output $\div$ total input $\leq 1$, always. If your calculation gives $\eta > 1$, you have either substituted into the wrong slot, mixed input with output, or used the wrong sign. Efficiencies are usually quoted as a percentage (×100). The "wasted" energy (almost always thermal) equals $E_{\text{input}}(1 - \eta)$.
What's NOT in this cheatsheet
This page gives you the formulas and the traps. The full Photon Academy A.3 Work, Energy and Power library (only available to enrolled students or via the resource library subscription) adds:
Notes PDF — every concept worked through in full, with derivations and exam-style commentary.
Tutorial booklet — IB-style questions sequenced from work-energy theorem through Sankey diagrams and energy density.
Tutorial Solutions — full mark-scheme-style worked solutions with M1/A1/B1 annotations.
Practice Solutions — extra past-paper-style problems with detailed walk-throughs.
Cheatsheet PDF — print-ready, brand-formatted, the same one our students take into mock exams.
