IB Physics flashcards · SL and HL · first assessment 2025

A.3 Work, energy and power flashcards: IB Physics SL and HL

Revision flashcards for A.3 Work, energy and power, written for the IB Diploma Physics course first assessed in 2025. Cards common to both levels are marked SL & HL; extension material is marked HL only. Use study mode to test yourself one card at a time, or read the full list below with every answer.

Work done, the work–energy principle, elastic energy, power, efficiency and energy density.

74 cards
Definition: 17
Equation: 12
Concept/explain: 17
SL and HL

Card type

Level

Order

All 74 A.3 Work, energy and power cards

DefinitionSL & HL
Define work done by a force.
Show answer
Work = force × displacement in the direction of the force: W = Fs cosθ. Scalar. SI unit: joule (1 J = 1 N m). Work is done only when there is a component of displacement along the force.
DefinitionSL & HL
Define kinetic energy.
Show answer
Energy an object possesses due to its motion: E_k = ½mv². Scalar, joule. Equal to the work needed to accelerate the mass from rest to speed v.
DefinitionSL & HL
Define gravitational potential energy (near Earth's surface).
Show answer
Energy stored due to an object's position in a gravitational field; change ΔE_p = mgΔh for small height changes where g is constant. Only CHANGES in E_p are physically meaningful; choose a reference level.
DefinitionSL & HL
Define elastic potential energy.
Show answer
Energy stored in a deformed elastic object (spring): E_H = ½kΔx² = area under the force–extension graph. Δx = extension or compression from natural length.
DefinitionSL & HL
State the principle of conservation of energy.
Show answer
Energy cannot be created or destroyed; it can only be transferred between forms/stores. The total energy of an isolated system is constant. Mechanical energy (E_k + E_p) is conserved only if no non-conservative forces (friction, drag) do work.
DefinitionSL & HL
Define power.
Show answer
Rate of doing work or transferring energy: P = ΔE/Δt = W/t. Scalar. SI unit: watt (1 W = 1 J s⁻¹).
DefinitionSL & HL
Define efficiency.
Show answer
Efficiency = useful work (energy) output ÷ total work (energy) input = useful power output ÷ total power input. Ratio (no unit) or percentage; always < 1 (100%) because some energy is dissipated, usually as thermal energy.
DefinitionSL & HLData booklet: No – memorise
State the work–energy theorem.
Show answer
The net work done on a body by all forces equals its change in kinetic energy: W_net = ΔE_k = ½mv² − ½mu². Not printed in booklet—memorise. Useful for variable forces where suvat fails.
EquationSL & HLData booklet: Yes
Write the equation for work done and explain the role of θ.
Show answer
W = Fs cosθ. F = force (N); s = displacement (m); θ = angle between force and displacement. θ = 0: W = Fs; θ = 90°: W = 0 (e.g. centripetal force, normal force on horizontal surface); θ = 180°: W = −Fs (friction). Misuse: using the angle to the wrong reference.
EquationSL & HLData booklet: Yes
Write kinetic energy in terms of velocity AND in terms of momentum.
Show answer
E_k = ½mv² = p²/(2m). p = momentum (kg m s⁻¹). Both in booklet. Useful: for equal momentum, lighter body has more KE (explosions/recoil). Misuse: forgetting to square v when doubling speed (KE ×4).
EquationSL & HLData booklet: Yes
Write the equation for change in gravitational potential energy and state its limitation.
Show answer
ΔE_p = mgΔh. m = mass (kg); g = 9.8 m s⁻²; Δh = change in height (m). Valid only where g is uniform (small Δh compared with Earth's radius). Misuse: using with satellites (must use E_p = −GMm/r, Theme D).
EquationSL & HLData booklet: Yes
Write the equation for elastic potential energy.
Show answer
E_H = ½kΔx². k = spring constant (N m⁻¹); Δx = extension/compression (m). Misuse: using E = kΔx² (forgetting ½) or E = ½F Δx with the wrong F (must be the FINAL force since force varies).
EquationSL & HLData booklet: Yes
Write the two equations for power and identify when each applies.
Show answer
P = ΔE/Δt (general); P = Fv (force F acting on object moving at velocity v in direction of F, constant velocity). Units W. Misuse: using P = Fv with the driving force when object is accelerating without accounting for resultant, or using average v for instantaneous P.
EquationSL & HLData booklet: Yes
Write the efficiency equation in energy and power forms.
Show answer
η = useful work out / total work in = useful power out / total power in. Dimensionless; multiply by 100 for %. Misuse: dividing by output, or using total output instead of USEFUL output.
Concept/explainSL & HL
Explain what happens to the kinetic energy of a block sliding to rest on a rough surface.
Show answer
- Friction acts opposite to motion
- Friction does negative work on block: W = −F_f s
- Kinetic energy decreases by this amount
- Energy transferred to thermal energy of block and surface (internal energy)
- Total energy conserved; mechanical energy not.
Concept/explainSL & HL
Explain why a centripetal force does no work on an object in uniform circular motion.
Show answer
- Force is always directed towards centre
- Velocity/displacement is tangential
- Force perpendicular to displacement (θ = 90°)
- W = Fs cos90° = 0
- Hence kinetic energy and speed remain constant.
Concept/explainSL & HL
Describe the energy changes in a simple pendulum swinging from one extreme to the other (no air resistance).
Show answer
- Extreme: maximum E_p, zero E_k (momentarily at rest)
- Descending: E_p → E_k
- Lowest point: maximum E_k, minimum E_p, maximum speed
- Rising: E_k → E_p
- Total mechanical energy constant
- Tension does no work (perpendicular to motion).
Concept/explainSL & HL
Explain why a car has a maximum speed on a level road, using P = Fv.
Show answer
- Engine delivers maximum power P
- Driving force F = P/v decreases as v increases
- Resistive forces (drag) increase with v
- Acceleration when F > resistive forces
- Maximum speed when driving force = total resistive force, so P = F_resistive × v_max.
Concept/explainSL & HL
Explain why a bouncing ball does not return to its release height.
Show answer
- At each bounce ball deforms; some KE → thermal energy and sound
- Air resistance also does negative work during flight
- Less mechanical energy after each bounce
- Lower E_p reachable, so lower maximum height
- Energy overall still conserved.
Concept/explainSL & HL
Explain how energy is transferred when a mass on a vertical spring is released from the unstretched position.
Show answer
- Initially E_p (gravitational) maximum; elastic energy zero
- Falling: gravitational E_p → E_k + elastic E_H
- Equilibrium position: E_k maximum
- Below equilibrium: E_k → elastic E_H
- Lowest point: E_k zero; loss of mg(2x_eq) = ½k(2x_eq)²
- Oscillation repeats if no damping.
Worked problemSL & HLData booklet: Yes
A 1200 kg car accelerates from rest to 25 m s⁻¹ in 10 s on a level road. Calculate the average useful power.
Show answer
ΔE_k = ½(1200)(25²) = 375 000 J. P = ΔE/Δt = 375 000/10 = 3.75 × 10⁴ W ≈ 38 kW (ignores resistive forces, so actual engine power is higher).
Worked problemSL & HLData booklet: Yes
A 2.0 kg ball is released from 5.0 m. Use energy to find its speed just before impact.
Show answer
mgh = ½mv² → v = √(2gh) = √(2 × 9.8 × 5.0) = √98 = 9.9 m s⁻¹. Independent of mass.
Worked problemSL & HLData booklet: Yes
A box is pushed 10 m along a floor by a 50 N force at 30° above the horizontal. Find the work done by the force.
Show answer
W = Fs cosθ = 50 × 10 × cos30° = 50 × 10 × 0.866 = 433 J ≈ 430 J.
Worked problemSL & HLData booklet: Yes
A spring (k = 200 N m⁻¹) compressed by 0.10 m launches a 0.050 kg ball horizontally. Find the launch speed.
Show answer
E_H = ½kΔx² = ½(200)(0.10²) = 1.0 J. ½mv² = 1.0 → v = √(2 × 1.0/0.050) = √40 = 6.3 m s⁻¹.
Worked problemSL & HLData booklet: Yes
A pump raises 2.0 kg of water per second through 10 m and ejects it at 5.0 m s⁻¹. Find the minimum power.
Show answer
Per second: ΔE_p = 2.0 × 9.8 × 10 = 196 J; E_k = ½(2.0)(5.0²) = 25 J. P = (196 + 25)/1 s = 221 W ≈ 220 W.
Worked problemSL & HLData booklet: Yes
A motor with 500 W input lifts a 20 kg load at a steady 1.5 m s⁻¹. Calculate its efficiency.
Show answer
Useful power = Fv = mgv = 20 × 9.8 × 1.5 = 294 W. η = 294/500 = 0.59 = 59%.
Worked problemSL & HLData booklet: Yes
A 70 kg cyclist travels at a constant 8.0 m s⁻¹ against a total resistive force of 40 N. Find the power developed.
Show answer
At constant velocity, driving force = resistive force = 40 N. P = Fv = 40 × 8.0 = 320 W.
Graph/diagramSL & HL
What does the area under a force–displacement graph represent?
Show answer
Work done by the force (energy transferred). For a spring (straight line through origin) area = ½kx² = elastic potential energy. For a variable force, count squares or integrate. Negative area = work done against the force.
Graph/diagramSL & HL
Describe graphs of E_k, E_p and total energy against height for an object in free fall from rest (no air resistance).
Show answer
E_p: straight line increasing with height (E_p = mgh). E_k: straight line decreasing with height, mirror image, zero at the top. Total energy: horizontal line (constant). The lines cross at half the drop height.
Graph/diagramSL & HL
Describe a Sankey diagram and its rules.
Show answer
Energy-flow diagram: arrow width ∝ energy (or power). Input on left; useful output continues straight; wasted energy (usually thermal) branches off (down). Total width conserved at every stage. Efficiency = useful width ÷ input width.
Exam technique/trapSL & HL
Trap: sign of work done by gravity and by friction.
Show answer
Gravity does POSITIVE work when object moves down (force and displacement same direction), NEGATIVE work when object rises. Friction and drag always do NEGATIVE work on the moving object (θ = 180°). Normal force on a horizontal surface does zero work.
Exam technique/trapSL & HLData booklet: Yes
Trap: unit slips with energy. Convert 1 kW h to joules.
Show answer
1 kW h = 1000 W × 3600 s = 3.6 × 10⁶ J = 3.6 MJ. Also: 1 eV = 1.60 × 10⁻¹⁹ J (Theme E); power in W, not J. Always check J = N m = kg m² s⁻².
Exam technique/trapSL & HL
Trap: when can you use energy conservation instead of suvat, and when must you not?
Show answer
Use energy methods for curved paths, springs (variable force), or when only speeds and heights are needed—energy is scalar, path-independent (if no friction). Must include work done against friction/drag if present: E_k,final = E_k,initial + ΔE_p − F_f s. Energy methods give no information about time or direction.
Exam technique/trapSL & HL
Practical: describe how to estimate your own power output climbing stairs and discuss the uncertainties.
Show answer
Measure mass m (bathroom scales), vertical height h (height of one step × number of steps), time t (stopwatch). P = mgh/t. Uncertainties: reaction time ≈ 0.2 s on t (use longer flights to reduce %), step heights not equal, ignores KE and internal work. Percentage uncertainty in P = sum of percentages in m, h, t.
DefinitionSL & HLData booklet: No – memorise
Define the specific energy of a fuel and state its unit.
Show answer
Specific energy E_SP is the energy released per unit MASS of fuel consumed: E_SP = energy released / mass, unit J kg⁻¹ (scalar). Exam tip: the mark is for "per unit mass", not "per kilogram of fuel burnt per second" (that is power). Do not write "energy density" — that term is reserved for the per-unit-volume quantity. Typical values expected: petrol ≈ 4.6 × 10⁷ J kg⁻¹, coal ≈ 3.0 × 10⁷ J kg⁻¹, hydrogen ≈ 1.4 × 10⁸ J kg⁻¹, nuclear fission of U-235 ≈ 7 × 10¹³ J kg⁻¹. Students lose the unit mark by writing J/kg s or MJ without conversion.
DefinitionSL & HLData booklet: No – memorise
Define the energy density of a fuel and state its unit.
Show answer
Energy density E_D is the energy released per unit VOLUME of fuel consumed: E_D = energy released / volume, unit J m⁻³ (scalar). Exam tip: the commonest error is quoting J kg⁻¹ — that is specific energy. E_D = E_SP × ρ, so a fuel with high specific energy but low density (gaseous hydrogen) has a poor energy density. Energy density decides tank/storage SIZE; specific energy decides the MASS you must carry. Both are properties of the fuel, not of the engine, so neither is affected by efficiency.
DefinitionSL & HL
Define mechanical energy and state the condition under which it is conserved.
Show answer
Mechanical energy = kinetic energy + potential energy (gravitational + elastic) of a system. It is conserved when no resultant work is done by dissipative (non-conservative) forces such as friction, air resistance or internal deformation. Exam tip: "in a vacuum" or "no friction" alone is not enough if a rope is being stretched inelastically or the surface is deforming — say "no energy is transferred to internal (thermal) energy". Mechanical energy is a scalar measured in joules; total energy is always conserved even when mechanical energy is not.
DefinitionSL & HL
Define a dissipative (non-conservative) force and give two examples.
Show answer
A dissipative force is one for which the work done in moving between two points depends on the path taken, and which transfers mechanical energy to internal (thermal) energy of the system and surroundings — it cannot be recovered as potential energy. Examples: kinetic friction, air/fluid drag, viscous damping. Exam tip: the mark is for path dependence OR the irreversible transfer to internal energy — a bare statement that "energy is lost" scores zero, because energy is never lost, only degraded. Work done by a dissipative force on a moving body is always negative.
DefinitionSL & HL
Define a conservative force and state the property that makes gravitational potential energy definable.
Show answer
A conservative force is one for which the work done between two points is independent of the path, and is zero around any closed loop; the work done can therefore be written as the loss of a potential energy function, W = −ΔE_p. Gravity and the ideal spring force are conservative. Exam tip: this path independence is exactly why E_p can be defined at all — friction has no associated potential energy for this reason. The mark scheme accepts "work done in a closed loop is zero" as a full definition; "energy is stored" alone is not enough.
DefinitionSL & HLData booklet: No – memorise
Define the spring constant k and state its unit and the limitation on its use.
Show answer
The spring constant (force constant) k is the force required per unit extension: k = F/Δx, unit N m⁻¹ (scalar). It equals the gradient of the linear region of a force–extension graph. Exam tip: k is only constant while the spring obeys Hooke's law, i.e. below the limit of proportionality; beyond it the graph curves and ½kΔx² is no longer valid — use the area under the graph instead. Common slips: giving the unit as N m or N m⁻², and using extension in cm without converting to m (this makes k a factor of 100 too small).
DefinitionSL & HLData booklet: No – memorise
Define the watt and express it in SI base units.
Show answer
One watt is one joule of energy transferred per second: 1 W = 1 J s⁻¹. In base units 1 W = 1 kg m² s⁻³ (since 1 J = 1 kg m² s⁻²). Exam tip: a "show that the unit of Fv is the watt" question wants N × m s⁻¹ = kg m s⁻² × m s⁻¹ = kg m² s⁻³ — show the substitution, do not just assert it. Power is a scalar. Watch the kW h trap: that is an energy unit (3.6 × 10⁶ J), not a power unit, and W h⁻¹ is meaningless in IB answers.
DefinitionSL & HL
Distinguish between average power and instantaneous power.
Show answer
Average power = total energy transferred / total time taken, P_av = ΔE/Δt. Instantaneous power is the rate of energy transfer at one instant, given by P = Fv with the instantaneous speed v (strictly the limit of ΔE/Δt as Δt → 0). Exam tip: for a body accelerating from rest under a constant force, the instantaneous power at the end is TWICE the average power over the journey (since v_av = ½v_max) — a very common trap in lift and car questions. Read the question: "at the moment when the speed is…" means instantaneous.
DefinitionSL & HL
Define degraded energy and explain its role in stating efficiency.
Show answer
Degraded energy is energy that has been transferred to the surroundings, usually as internal (thermal) energy at low temperature, in a form that is too dispersed to do useful work. Efficiency = useful output / total input, so degraded energy is exactly the fraction (1 − efficiency) of the input. Exam tip: the mark scheme will not accept "energy is lost as heat" — write "transferred to internal energy of the surroundings" or "degraded". Efficiency has no unit and is ≤ 1 (or ≤ 100%); quoting an efficiency greater than 1 always indicates that useful and total inputs have been swapped.
EquationSL & HLData booklet: No – derive
Write an expression for the work done in stretching an ideal spring from extension x₁ to extension x₂.
Show answer
W = ½kx₂² − ½kx₁², where W = work done (J), k = spring constant (N m⁻¹), x = extension measured from the natural length (m). This is the difference of the areas under the F–x graph. Conditions: spring obeys Hooke's law throughout, extensions measured from the UNSTRETCHED length. Data booklet: E_H = ½kΔx² is printed; the difference form must be derived. Common misuse: using ½k(x₂ − x₁)² — wrong, because the force is not zero at x₁. Sanity check: k = 200 N m⁻¹, 0.10 m → 0.20 m gives ½(200)(0.04 − 0.01) = 3.0 J, not ½(200)(0.01) = 1.0 J.
EquationSL & HLData booklet: Yes
Write the general expression for the power delivered by a force whose direction differs from the velocity.
Show answer
P = Fv cosθ, where P = instantaneous power (W), F = magnitude of the force (N), v = instantaneous speed (m s⁻¹), θ = angle between force and velocity vectors. Conditions: instantaneous values; reduces to P = Fv when θ = 0. Data booklet: P = Fv is printed, the cosθ form follows from W = Fs cosθ divided by Δt. Common misuse: applying P = Fv to a centripetal force — there θ = 90°, cos 90° = 0, so P = 0 and no work is done. Sanity check: a 100 N tow rope at 30° to the motion at 5.0 m s⁻¹ delivers 100 × 5.0 × 0.866 = 4.3 × 10² W.
EquationSL & HLData booklet: No – memorise
Write the relationship linking energy released, specific energy, energy density and density of a fuel.
Show answer
Energy released E = m × E_SP = V × E_D, and E_D = E_SP × ρ. Symbols: E in J, m = mass burnt (kg), V = volume burnt (m³), E_SP in J kg⁻¹, E_D in J m⁻³, ρ = density in kg m⁻³. Conditions: complete combustion, values quoted for the fuel alone (oxidiser mass excluded). Data booklet: not printed — memorise. Common misuse: multiplying specific energy by volume. Sanity check: petrol, E_SP = 4.6 × 10⁷ J kg⁻¹, ρ = 740 kg m⁻³ → E_D = 3.4 × 10¹⁰ J m⁻³.
EquationSL & HLData booklet: No – derive
Write an expression for the energy dissipated by kinetic friction and state its sign convention.
Show answer
Energy transferred to internal energy ΔE_int = f d = μ_d F_N d, where f = friction force (N), d = distance moved along the surface (m), μ_d = coefficient of dynamic friction (no unit), F_N = normal force (N). The WORK done by friction on the body is W_f = −f d (negative, since friction opposes the displacement). Conditions: constant friction force; d is path length, not displacement. Data booklet: F_f ≤ μ_d F_N appears in A.2. Common misuse: using the straight-line displacement for a curved or there-and-back path — friction dissipates on every metre travelled, so a round trip dissipates 2f d while gravity does zero net work.
EquationSL & HLData booklet: Yes
Write an expression for the change in gravitational potential energy of a body moving along an incline of length L at angle θ.
Show answer
ΔE_p = mgΔh = mgL sinθ, where m = mass (kg), g = 9.8 m s⁻² (gravitational field strength, N kg⁻¹), L = distance along the slope (m), θ = angle of the slope to the horizontal, Δh = vertical height gained (m). Conditions: uniform field, so only near the Earth's surface. Data booklet: ΔE_p = mgΔh is printed; the L sinθ substitution must be done by you. Common misuse: using L instead of L sinθ, or cosθ instead of sinθ. Sanity check: 5.0 kg up a 4.0 m slope at 30° gives 5.0 × 9.8 × 4.0 × 0.50 = 98 J.
EquationSL & HLData booklet: Yes
State how the kinetic energy of a body changes when its speed changes by a given factor, and justify it.
Show answer
Since E_k = ½mv², E_k ∝ v² at constant mass, so multiplying the speed by n multiplies the kinetic energy by n². Symbols: E_k (J), m (kg), v (m s⁻¹). Conditions: v much less than c (non-relativistic), constant mass. Data booklet: E_k = ½mv² and E_k = p²/2m are printed. Common misuse: assuming doubling the speed doubles the braking distance — it quadruples it, because the constant braking force must remove four times the energy over F d. Sanity check: 10 → 30 m s⁻¹ is ×3 in speed and ×9 in E_k, so nine times the stopping distance.
Concept/explainSL & HL
Explain why the work done against friction depends on the path taken while the work done against gravity does not.
Show answer
- Gravity is a conservative force: the work done depends only on the vertical height difference Δh, so W = −mgΔh for any path
- Around a closed loop gravity does zero net work, which is why a gravitational potential energy function can be defined
- Friction always acts opposite to the instantaneous velocity, so its work contribution is negative on every element of the path
- The total dissipated energy is therefore f × (total path length), which grows with a longer or more winding route
- A body returning to its starting point regains all its E_p but has permanently transferred f × 2d to internal energy
- Hence no "friction potential energy" exists. Exam tip: incomplete answers say friction "loses energy" without identifying the transfer to internal energy of the surfaces and surroundings.
Concept/explainSL & HL
Explain, using energy, why the speed of a block at the bottom of a smooth ramp is independent of the ramp's angle and shape.
Show answer
- On a smooth ramp the normal force is always perpendicular to the velocity, so it does no work
- The only force doing work is gravity, which is conservative
- Energy conservation gives ½mv² = mgΔh, so v = √(2gΔh)
- Δh is fixed by the start and end heights, so the path taken between them is irrelevant
- The mass cancels, so all masses arrive with the same speed
- The angle changes the TIME taken and the acceleration, but not the final speed. Exam tip: students often try suvat with a fixed a = g sinθ and fail on a curved ramp; the energy method is valid for any shape. If friction acts, the answer does depend on shape because the path length changes.
Concept/explainSL & HL
Explain why the power output of a lift motor is greater while the lift accelerates upwards than when it moves upwards at constant speed at the same instantaneous speed.
Show answer
- At constant velocity the tension equals the weight, T = mg, so P = mgv
- While accelerating upwards the resultant force is upwards, so T = m(g + a) > mg
- The motor must supply energy for the gain in kinetic energy as well as the gain in gravitational potential energy
- Therefore at the same instantaneous speed P = m(g + a)v is larger
- On deceleration T = m(g − a) and the required power falls; a counterweight further reduces the effective mass lifted
- Real motors must also supply the energy dissipated in friction and in the cables. Exam tip: many students use P = mgv throughout and ignore the acceleration term entirely.
Concept/explainSL & HL
Describe the energy transfers for a skydiver from release until after the parachute opens at terminal velocity.
Show answer
- Immediately after release the only significant force is weight, so nearly all the E_p lost becomes E_k and the diver accelerates at ≈ g
- As speed rises the drag force grows, so a growing fraction of the E_p lost is transferred to internal energy of the air and the diver
- At terminal velocity the resultant force is zero, so E_k is constant and ALL the E_p lost per second is dissipated: mgv = F_drag v
- Opening the parachute increases drag sharply, so E_k falls rapidly and a large amount of energy is dissipated in a short time
- A new, lower terminal velocity is reached where mg = F_drag again. Exam tip: at terminal velocity the diver still loses E_p — a common error is claiming "no energy change" because the speed is constant.
Concept/explainSL & HL
Explain why a fuel's specific energy determines the mass carried while its energy density determines the storage volume, using hydrogen and petrol as the example.
Show answer
- Specific energy E_SP is energy per kilogram, so the fuel MASS needed for a required energy is m = E/E_SP
- Energy density E_D is energy per cubic metre, so the fuel VOLUME needed is V = E/E_D
- Hydrogen has a very high E_SP (≈ 1.4 × 10⁸ J kg⁻¹, about three times petrol) so it is attractive where mass matters, e.g. rockets
- But as a gas at atmospheric pressure hydrogen has a very low density, so its E_D is orders of magnitude below petrol's ≈ 3.4 × 10¹⁰ J m⁻³
- Hence it must be compressed or liquefied, which costs energy and needs heavy tanks
- E_D = E_SP × ρ links the two. Exam tip: answers that treat the two terms as synonyms lose both marks.
Concept/explainSL & HL
Compare the specific energies of coal, petrol, hydrogen and nuclear fuel and account for the enormous difference of the last.
Show answer
- Coal ≈ 3 × 10⁷ J kg⁻¹, petrol ≈ 4.6 × 10⁷ J kg⁻¹, hydrogen ≈ 1.4 × 10⁸ J kg⁻¹, U-235 fission ≈ 7 × 10¹³ J kg⁻¹
- Chemical fuels release energy by rearranging electrons in molecular bonds, involving energies of a few eV per molecule
- Nuclear fuels release energy from the strong nuclear interaction via mass defect, of order MeV per nucleus — about a million times greater per particle
- Hence nuclear specific energy exceeds chemical by roughly 10⁶
- This is why a nuclear plant refuels annually while a coal plant consumes trainloads daily
- Fuel choice also depends on E_D, cost, emissions and safety, not specific energy alone. Exam tip: quote at least one number with a correct unit — vague "much bigger" answers score little.
Concept/explainSL & HL
Explain why the efficiency of any real energy transfer is less than 100% and identify the Nature of Science idea involved.
Show answer
- Every real transfer involves dissipative forces or resistive heating, so some input energy is transferred to internal energy of the surroundings
- That energy is degraded: it is spread over many particles at low temperature and cannot all be recovered as useful work
- Energy is still conserved — the principle of conservation of energy is never broken; only the USEFUL fraction falls
- Efficiency = useful output / total input < 1
- Nature of Science: conservation of energy is a universal, repeatedly tested law, and every claimed "over-unity" or perpetual-motion machine has failed replication, which is why such claims are rejected on principle
- Efficiency is a ratio, so it has no unit. Exam tip: never write "energy is lost"; write "transferred to the surroundings as internal energy".
Concept/explainSL & HL
Explain the energy transfers in an Atwood machine (two masses connected over a smooth pulley) and why the string tension does no net work.
Show answer
- The heavier mass m₁ falls a distance d, losing m₁gd of gravitational potential energy
- The lighter mass m₂ rises the same distance d, gaining m₂gd
- The net loss (m₁ − m₂)gd becomes the kinetic energy of BOTH masses: (m₁ − m₂)gd = ½(m₁ + m₂)v²
- The string is inextensible and the pulley smooth, so both masses have the same speed
- Tension does positive work +Td on the rising mass and negative work −Td on the falling one, so the net work by tension is zero — it merely transfers energy between the two bodies
- A massive pulley would also gain rotational kinetic energy, reducing v. Exam tip: forgetting that both masses share the kinetic energy is the standard error.
Concept/explainSL & HL
Explain why, in an explosion in which two fragments of unequal mass separate, the lighter fragment carries away more kinetic energy.
Show answer
- Momentum is conserved and the total is zero, so the fragments carry equal and opposite momenta of magnitude p
- Kinetic energy is E_k = p²/2m, so for the same p, E_k ∝ 1/m
- Hence the smaller mass carries the larger share of the released energy
- The energy released comes from chemical or nuclear potential energy stored in the system, so total kinetic energy INCREASES — the collision is superelastic
- The energy split is m₂/(m₁ + m₂) for fragment 1
- The same reasoning explains recoil: a rifle carries far less kinetic energy than its bullet. Exam tip: students often assume equal energies because the momenta are equal — state E_k = p²/2m explicitly to earn the mark.
Concept/explainSL & HL
Explain why energy must be supplied continuously to keep a vehicle moving at constant speed even though its kinetic energy is not changing.
Show answer
- At constant speed the resultant force is zero, so by the work–energy theorem the NET work done on the vehicle is zero
- This does not mean no work is done: the driving force does positive work while drag and friction do equal negative work
- The engine must therefore supply energy at the rate P = Fv just to replace the energy dissipated to internal energy of the air, tyres and road
- On a hill the engine must additionally supply mgv sinθ for the rate of gain of gravitational potential energy
- Stop supplying energy and the resistive forces decelerate the vehicle. Exam tip: "the kinetic energy is constant so no energy is needed" is a classic zero-mark answer; identify the continuous dissipation.
Concept/explainSL & HL
Explain how the work done by a force that varies with displacement is found, and why W = Fs cosθ cannot be used directly.
Show answer
- W = Fs cosθ assumes F is constant in magnitude and direction over the whole displacement
- If F varies, split the displacement into elements Δs so small that F is effectively constant over each
- The work in each element is FΔs cosθ, and the total is the sum of all elements
- In the limit this sum is the area under the force–displacement graph (using the component of F along the displacement)
- For a spring this gives the triangular area ½FΔx = ½kΔx²
- Areas below the axis count as negative work
- Where no graph is given, count squares or use the mean force only if the graph is linear. Exam tip: using the maximum force rather than the mean force for a linear spring doubles the answer.
Worked problemSL & HL
A 5.0 kg block starts from rest and slides 4.0 m down a 30° incline for which μ_d = 0.20. Calculate its speed at the bottom.
Show answer
Use conservation of energy with dissipation: mgL sinθ = ½mv² + μ_d mg cosθ × L. Gain from gravity: 5.0 × 9.8 × 4.0 × sin30° = 98 J. Normal force F_N = mg cosθ = 5.0 × 9.8 × cos30° = 42.4 N, so friction f = 0.20 × 42.4 = 8.49 N and energy dissipated = 8.49 × 4.0 = 34 J. Kinetic energy at the bottom = 98 − 34 = 64 J. v = √(2 × 64 / 5.0) = √25.6 = 5.1 m s⁻¹ (2 s.f.). Check/Trap: use cosθ for the normal force and sinθ for the height — swapping them is the standard error; the mass does not cancel here only because friction also scales with m, so in fact it does cancel: v = √(2gL(sinθ − μ_d cosθ)).
Worked problemSL & HL
A lift of total mass 800 kg is accelerating upwards at 1.2 m s⁻² at the instant its speed is 3.0 m s⁻¹. Calculate the power delivered by the motor at that instant.
Show answer
Newton's second law on the lift: T − mg = ma, so T = m(g + a) = 800 × (9.8 + 1.2) = 800 × 11.0 = 8800 N. Instantaneous power P = Tv = 8800 × 3.0 = 2.64 × 10⁴ W ≈ 2.6 × 10⁴ W (2 s.f.). Check/Trap: using P = mgv gives only 2.35 × 10⁴ W and ignores the rate of gain of kinetic energy — always find the tension first. If the lift were decelerating at 1.2 m s⁻² the tension would be 800 × 8.6 = 6880 N and P = 2.1 × 10⁴ W. Friction and a counterweight would change T further, so state any assumption that the cable is light and the pulley smooth.
Worked problemSL & HL
Petrol has a specific energy of 4.6 × 10⁷ J kg⁻¹ and a density of 740 kg m⁻³. Calculate its energy density and the energy available from a 50 litre tank.
Show answer
Energy density E_D = E_SP × ρ = 4.6 × 10⁷ × 740 = 3.404 × 10¹⁰ ≈ 3.4 × 10¹⁰ J m⁻³ (2 s.f.). Volume: 50 litres = 50 × 10⁻³ m³ = 0.050 m³. Energy E = E_D × V = 3.404 × 10¹⁰ × 0.050 = 1.702 × 10⁹ ≈ 1.7 × 10⁹ J. Equivalently m = ρV = 740 × 0.050 = 37 kg and E = 37 × 4.6 × 10⁷ = 1.7 × 10⁹ J — the two routes must agree. Check/Trap: 1 litre = 10⁻³ m³, not 10⁻² m³; forgetting this is worth a factor of ten. If the engine is 25% efficient the useful output is only 4.3 × 10⁸ J.
Worked problemSL & HL
Water falls at 500 kg s⁻¹ through a vertical drop of 60 m onto a turbine of overall efficiency 80%. Calculate the electrical power output.
Show answer
Rate of loss of gravitational potential energy = (m/t)gΔh = 500 × 9.8 × 60 = 2.94 × 10⁵ W. This is the input power. Useful output = efficiency × input = 0.80 × 2.94 × 10⁵ = 2.35 × 10⁵ ≈ 2.4 × 10⁵ W (2 s.f.). Check/Trap: 500 kg s⁻¹ is a mass FLOW rate, so the mgΔh calculation already gives a power — do not divide by a time again. If the water still leaves the turbine at 4.0 m s⁻¹, that carries away ½ × 500 × 4.0² = 4.0 × 10³ W of kinetic energy per second, part of the 20% not converted.
Worked problemSL & HL
A stationary object explodes into fragments of 2.0 kg and 3.0 kg, releasing 200 J of kinetic energy. Calculate the speed of each fragment.
Show answer
Momentum is conserved and the initial momentum is zero, so the fragments have equal and opposite momenta of magnitude p. Using E_k = p²/2m: total E_k = p²/(2 × 2.0) + p²/(2 × 3.0) = p²(0.250 + 0.1667) = 0.4167p² = 200 J, so p² = 480 and p = 21.9 kg m s⁻¹. Then v₂ = p/m = 21.9/2.0 = 11 m s⁻¹ and v₃ = 21.9/3.0 = 7.3 m s⁻¹ (2 s.f.). Check: ½ × 2.0 × 11.0² = 121 J and ½ × 3.0 × 7.30² = 80 J, total 201 J ✓. Check/Trap: the fragments do NOT share the energy equally — the lighter one takes 60% because E_k ∝ 1/m at equal momentum.
Worked problemSL & HL
A 10 g bullet travelling at 400 m s⁻¹ embeds in a 2.0 kg block free to move on a frictionless surface. Calculate the kinetic energy transferred to internal energy.
Show answer
Momentum is conserved in the collision: (0.010 × 400) = (0.010 + 2.0)v, so v = 4.00/2.010 = 1.99 m s⁻¹. Initial E_k = ½ × 0.010 × 400² = 800 J. Final E_k = ½ × 2.010 × 1.99² = 3.98 J. Energy transferred to internal energy = 800 − 3.98 = 796 ≈ 8.0 × 10² J, i.e. 99.5% of the original kinetic energy. Check/Trap: kinetic energy is NOT conserved in a perfectly inelastic collision — never start from ½mv² to find v; always use momentum first. Include the bullet's mass in the final combined mass; omitting it changes v by 0.5%.
Worked problemSL & HL
A 0.20 kg ball is dropped from 1.5 m above the top of a vertical spring of constant 800 N m⁻¹. Calculate the maximum compression of the spring.
Show answer
At maximum compression x the ball is momentarily at rest, so all the gravitational potential energy lost becomes elastic potential energy: mg(h + x) = ½kx². Note the ball falls h + x in total. 0.20 × 9.8 × (1.5 + x) = ½ × 800 × x², i.e. 1.96(1.5 + x) = 400x², giving 400x² − 1.96x − 2.94 = 0. Solving: x = [1.96 + √(3.84 + 4704)]/800 = 70.6/800 = 0.088 m (8.8 cm, 2 s.f.). Check: ½ × 800 × 0.088² = 3.1 J and mg(1.588) = 3.1 J ✓. Check/Trap: omitting the extra fall x gives 0.086 m and loses the method mark; take the positive root only.
Graph/diagramSL & HL
Describe the force–extension graph for a spring loaded beyond its limit of proportionality, including what the gradient and area mean.
Show answer
Axes: force F/N on the y-axis against extension Δx/m on the x-axis. The graph is a straight line through the origin up to the limit of proportionality, then curves with decreasing gradient. Gradient of the straight section = spring constant k (N m⁻¹). Area under the whole graph = work done in stretching = elastic potential energy stored, so ½kΔx² is valid only in the linear region; beyond it, count squares or use the trapezium rule. If the elastic limit is exceeded, the unloading line is parallel but displaced, leaving a permanent extension, and the area between loading and unloading curves is the energy dissipated as internal energy. A stiffer spring gives a steeper line and stores more energy for the same extension.
Graph/diagramSL & HL
Describe a graph of energy transferred against time for a motor, and explain how to obtain both average and instantaneous power from it.
Show answer
Axes: energy transferred E/J on the y-axis against time t/s on the x-axis. The gradient at a point is the instantaneous power, P = dE/dt; the gradient of the chord between two points is the average power over that interval. A straight line means constant power; a curve of increasing gradient means the power output is rising (for example a motor lifting a load that is speeding up). The intercept is the energy already transferred at t = 0 and should normally be zero. To find instantaneous power at a chosen time, draw a tangent and take Δy/Δx with units J/s = W. If the load doubles at constant speed, the line's gradient doubles.
Graph/diagramSL & HL
Describe graphs of kinetic energy against speed and against momentum, and explain how to linearise the first to find the mass.
Show answer
E_k against v: a parabola through the origin, since E_k = ½mv², curving upwards — the gradient is not the mass and increases with v. E_k against p: also a parabola through the origin, since E_k = p²/2m. To linearise, plot E_k on the y-axis against v² on the x-axis: this gives a straight line through the origin of gradient ½m, so m = 2 × gradient. Equivalently plot E_k against p² for gradient 1/2m. A non-zero positive intercept on the E_k–v² line suggests a systematic error such as a timing offset in the light gates, while scatter about the line indicates random error. Error bars in v² are twice the fractional uncertainty in v.
Exam technique/trapSL & HL
Trap: identifying the correct distance to use in W = Fs cosθ when a pulley system or a moving point of application is involved.
Show answer
The trap: s is the displacement of the POINT OF APPLICATION of the force, resolved along the force, not the displacement of the load. In a pulley system with two supporting rope sections, the free end of the rope moves twice as far as the load rises, so the effort force is halved but the work done is unchanged (ideally). Students fall for it because they read "the load rises 2.0 m" and use s = 2.0 m for the effort. Correct approach: identify which force you are asked about, find how far ITS point of application moves, then apply W = Fs cosθ. Command terms: "calculate" needs working and a unit; "show that" needs one more significant figure than the printed value and a final statement.
Exam technique/trapSL & HL
Trap: propagating uncertainties into a kinetic energy or power calculation.
Show answer
The trap: uncertainties in squared quantities must be DOUBLED as fractional uncertainties. For E_k = ½mv², the fractional uncertainty is Δm/m + 2Δv/v. Students fall for it by adding the raw absolute uncertainties or by forgetting the factor of two on v. Worked example: m = 2.0 ± 0.1 kg (5%) and v = 12.0 ± 0.4 m s⁻¹ (3.3%) give E_k = 144 J with fractional uncertainty 5% + 6.7% = 11.7% ≈ 12%, so E_k = 144 ± 17 J, quoted as (1.4 ± 0.2) × 10² J. For P = ΔE/Δt the fractional uncertainties add. Round the uncertainty to one significant figure and match the value's decimal place; never quote more significant figures than the raw data justify.
Exam technique/trapSL & HL
Practical: describe an investigation of how the efficiency of a small electric motor varies with the load it lifts.
Show answer
Apparatus: low-voltage motor with a spool, slotted masses on a thread, metre rule, light gates or video at 30 fps, ammeter and voltmeter (or joulemeter), clamp stand. Method: for each load m, measure the steady lifting speed v over a measured height, record V and I, then efficiency = mgv/(VI) — repeat three times and mean. Independent variable m; dependent variable efficiency; controls: same supply voltage, same motor, same height, motor allowed to reach steady speed before timing. Expect efficiency to rise, peak, then fall as the motor stalls. Limitations: the thread's mass and spool friction, motor heating changing coil resistance, reaction-time error if hand-timed. Improvements: use light gates, take readings only in the constant-velocity region, and plot efficiency against load with error bars.
Exam technique/trapSL & HL
Trap: choosing between the work–energy theorem and a full energy-conservation equation when dissipative forces act.
Show answer
The trap: writing ½mv² = mgΔh on a rough slope, or applying the work–energy theorem with only one of the forces. Students fall for it because the smooth-case equation is memorised as a formula rather than as a special case. Correct approach: the work–energy theorem uses the RESULTANT force — W_net = ΔE_k — so every force doing work must be included, including friction (negative) and any applied force. Alternatively write E_k initial + E_p initial + W_applied = E_k final + E_p final + energy dissipated. Command-term guidance: "estimate" permits reasonable assumptions, stated explicitly (for example "assume friction is negligible"); "determine" expects a numerical answer from given or graphical data with a unit and sensible significant figures.

Know it, then write it the way it is marked

One-to-one IB tuition that turns correct physics and maths into full-mark answers.

Book a free consultation

These flashcards are original ExaminerPrep material, written independently. They are not IB documents and do not reproduce IB syllabus text, examination papers or mark schemes; topic references follow the published subject guides. ExaminerPrep is an independent tutoring service. It has been developed independently from and is not endorsed by the International Baccalaureate Organization. "International Baccalaureate", "IB" and "IB Diploma Programme" are registered trademarks of the IBO, used here for descriptive purposes only.

All 74 A.3 Work, energy and power cards

Define work done by a force.

Define kinetic energy.

Define gravitational potential energy (near Earth's surface).

Define elastic potential energy.

State the principle of conservation of energy.

Define power.

Define efficiency.

State the work–energy theorem.

Write the equation for work done and explain the role of θ.

Write kinetic energy in terms of velocity AND in terms of momentum.

Write the equation for change in gravitational potential energy and state its limitation.

Write the equation for elastic potential energy.

Write the two equations for power and identify when each applies.

Write the efficiency equation in energy and power forms.

Explain what happens to the kinetic energy of a block sliding to rest on a rough surface.

Explain why a centripetal force does no work on an object in uniform circular motion.

Describe the energy changes in a simple pendulum swinging from one extreme to the other (no air resistance).

Explain why a car has a maximum speed on a level road, using P = Fv.

Explain why a bouncing ball does not return to its release height.

Explain how energy is transferred when a mass on a vertical spring is released from the unstretched position.

A 1200 kg car accelerates from rest to 25 m s⁻¹ in 10 s on a level road. Calculate the average useful power.

A 2.0 kg ball is released from 5.0 m. Use energy to find its speed just before impact.

A box is pushed 10 m along a floor by a 50 N force at 30° above the horizontal. Find the work done by the force.

A spring (k = 200 N m⁻¹) compressed by 0.10 m launches a 0.050 kg ball horizontally. Find the launch speed.

A pump raises 2.0 kg of water per second through 10 m and ejects it at 5.0 m s⁻¹. Find the minimum power.

A motor with 500 W input lifts a 20 kg load at a steady 1.5 m s⁻¹. Calculate its efficiency.

A 70 kg cyclist travels at a constant 8.0 m s⁻¹ against a total resistive force of 40 N. Find the power developed.

What does the area under a force–displacement graph represent?

Describe graphs of E_k, E_p and total energy against height for an object in free fall from rest (no air resistance).

Describe a Sankey diagram and its rules.

Trap: sign of work done by gravity and by friction.

Trap: unit slips with energy. Convert 1 kW h to joules.

Trap: when can you use energy conservation instead of suvat, and when must you not?

Practical: describe how to estimate your own power output climbing stairs and discuss the uncertainties.

Define the specific energy of a fuel and state its unit.

Define the energy density of a fuel and state its unit.

Define mechanical energy and state the condition under which it is conserved.

Define a dissipative (non-conservative) force and give two examples.

Define a conservative force and state the property that makes gravitational potential energy definable.

Define the spring constant k and state its unit and the limitation on its use.

Define the watt and express it in SI base units.

Distinguish between average power and instantaneous power.

Define degraded energy and explain its role in stating efficiency.

Write an expression for the work done in stretching an ideal spring from extension x₁ to extension x₂.

Write the general expression for the power delivered by a force whose direction differs from the velocity.

Write the relationship linking energy released, specific energy, energy density and density of a fuel.

Write an expression for the energy dissipated by kinetic friction and state its sign convention.

Write an expression for the change in gravitational potential energy of a body moving along an incline of length L at angle θ.

State how the kinetic energy of a body changes when its speed changes by a given factor, and justify it.

Explain why the work done against friction depends on the path taken while the work done against gravity does not.

Explain, using energy, why the speed of a block at the bottom of a smooth ramp is independent of the ramp's angle and shape.

Explain why the power output of a lift motor is greater while the lift accelerates upwards than when it moves upwards at constant speed at the same instantaneous speed.

Describe the energy transfers for a skydiver from release until after the parachute opens at terminal velocity.

Explain why a fuel's specific energy determines the mass carried while its energy density determines the storage volume, using hydrogen and petrol as the example.

Compare the specific energies of coal, petrol, hydrogen and nuclear fuel and account for the enormous difference of the last.

Explain why the efficiency of any real energy transfer is less than 100% and identify the Nature of Science idea involved.

Explain the energy transfers in an Atwood machine (two masses connected over a smooth pulley) and why the string tension does no net work.

Explain why, in an explosion in which two fragments of unequal mass separate, the lighter fragment carries away more kinetic energy.

Explain why energy must be supplied continuously to keep a vehicle moving at constant speed even though its kinetic energy is not changing.

Explain how the work done by a force that varies with displacement is found, and why W = Fs cosθ cannot be used directly.

A 5.0 kg block starts from rest and slides 4.0 m down a 30° incline for which μ_d = 0.20. Calculate its speed at the bottom.

A lift of total mass 800 kg is accelerating upwards at 1.2 m s⁻² at the instant its speed is 3.0 m s⁻¹. Calculate the power delivered by the motor at that instant.

Petrol has a specific energy of 4.6 × 10⁷ J kg⁻¹ and a density of 740 kg m⁻³. Calculate its energy density and the energy available from a 50 litre tank.

Water falls at 500 kg s⁻¹ through a vertical drop of 60 m onto a turbine of overall efficiency 80%. Calculate the electrical power output.

A stationary object explodes into fragments of 2.0 kg and 3.0 kg, releasing 200 J of kinetic energy. Calculate the speed of each fragment.

A 10 g bullet travelling at 400 m s⁻¹ embeds in a 2.0 kg block free to move on a frictionless surface. Calculate the kinetic energy transferred to internal energy.

A 0.20 kg ball is dropped from 1.5 m above the top of a vertical spring of constant 800 N m⁻¹. Calculate the maximum compression of the spring.

Describe the force–extension graph for a spring loaded beyond its limit of proportionality, including what the gradient and area mean.

Describe a graph of energy transferred against time for a motor, and explain how to obtain both average and instantaneous power from it.

Describe graphs of kinetic energy against speed and against momentum, and explain how to linearise the first to find the mass.

Trap: identifying the correct distance to use in W = Fs cosθ when a pulley system or a moving point of application is involved.

Trap: propagating uncertainties into a kinetic energy or power calculation.

Practical: describe an investigation of how the efficiency of a small electric motor varies with the load it lifts.

Trap: choosing between the work–energy theorem and a full energy-conservation equation when dissipative forces act.

Know it, then write it the way it is marked