Summary to follow. 11 syllabus statements · 24 questions · about twenty minutes.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank ·
Specialist review in progress
· How these pages are made
Assessed in Paper 1A (multiple choice), Paper 1B (data-based) and Paper 2 (short and extended response). IB Physics guide (first assessment 2025, updated November 2023).
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In preparation: 0 of 11 sections compiled and reviewed. The rest show key terms and common misconceptions from our question bank until they are.
Principle of the conservation of energy
Principle of the conservation of energy
Energy cannot be created or destroyed; it can only be transferred from one store to another or from one place to another. The total energy of an isolated system (one that exchanges no energy with its surroundings) therefore stays constant. The principle applies to the total energy of the whole system, not to any single object or any single store within it.
Dissipated energy
Energy transferred to the internal energy of a device and its surroundings, usually by heating and by sound, in a way that spreads it out so it can no longer be used to do useful work. Dissipated energy has not been destroyed: it is still present, so the total energy is unchanged, but it is no longer available for the intended purpose.
Students often think Energy is consumed when things happen, so after a process some of the energy no longer exists. In fact No. Energy is never used up or destroyed; it is transferred to other stores, often to internal energy of the surroundings, where it is spread out and less useful.
Students often think Energy and force are the same kind of thing, so energy can be turned into a force such as friction or the push of the floor. In fact No. Force (in N) is an interaction that can transfer energy by doing work; energy (in J) is a conserved quantity. Energy is never converted into a force.
Work done
Work done
A scalar quantity equal to the energy transferred by a force when its point of application moves. Doing work on a body is one way of transferring energy to (or from) it: a force that does 50 J of work on a body transfers 50 J of energy. If the point of application does not move, no work is done, however large the force. SI unit: joule (J).
Joule (J)
The SI unit of work and of energy. One joule is the work done when a force of 1 N moves its point of application through 1 m in the direction of the force, so 1 J = 1 N m = 1 kg m² s⁻². Work and energy share this unit because work done is a transfer of energy.
Students often think Anything that takes effort is work, so holding a heavy object still is doing a lot of work on it. In fact No. Work is done on an object only when a force moves its point of application. Holding an object stationary transfers no energy to it, so no work is done on it.
Students often think The work done by a force is the force multiplied by the time for which it acts, so a force that acts for longer always does more work. In fact No. Work depends on the force and the displacement along its line, W = Fs cos θ. Force multiplied by time is impulse, which equals the change in momentum; it is not a measure of the energy transferred.
Sankey diagram
Sankey diagram
A flow diagram of the energy transfers in a device or system. One arrow represents the total energy (or power) input and splits into arrows for each output, useful and wasted. The WIDTH of each arrow is proportional to the energy or power it represents, drawn to a stated scale. Because energy is conserved, the widths of the output arrows add up to the width of the input arrow. The efficiency is the width of the useful output arrow divided by the width of the input arrow.
Students often think In a Sankey diagram the longest arrow represents the largest energy transfer, as in a bar chart. In fact No. It is the width of each arrow, drawn to scale, that represents the energy or power; arrow lengths carry no meaning.
Students often think The width in millimetres, or the value on an arrow, can be read straight off as the answer, whatever the scale and whatever the input. In fact Only if the diagram's scale makes them so. Widths must be converted with the stated scale, and energies are percentages of the input only when the input is 100 units.
Work done by a constant force at an angle
Work done by a constant force at an angle
When a constant force F acts on a body whose displacement is s, and θ is the angle between the force and the displacement, the work done by the force is W = Fs cos θ. Only the component of the force along the line of displacement, F cos θ, does work. W is positive when θ < 90° (energy is transferred to the body), zero when θ = 90° (for example, the normal force on a body sliding along a level surface, or the centripetal force on a body in uniform circular motion), and negative when 90° < θ ≤ 180° (energy is transferred from the body, as by friction).
Students often think Work is force multiplied by distance moved, whatever the direction of the force relative to the motion. In fact No. Only the component of the force along the displacement does work: W = Fs cos θ. A force perpendicular to the displacement does no work.
Students often think The component of the force along the displacement is F sin θ, or sine and cosine are interchangeable. In fact The cosine of the angle between the force and the displacement: the component along the displacement is F cos θ.
Work done by the resultant force
Work done by the resultant force
The work done by the resultant (net) force on a system equals the change in the energy of the system. For a single body with gravity included as one of the forces, the work done by the resultant force equals the change in its kinetic energy: F_net s cos θ = ΔE_k. If the resultant force is zero, as at constant velocity, the resultant force does no work and the kinetic energy is unchanged, even though individual forces may each do work.
Students often think A moving body needs a resultant force in its direction of motion to keep it moving; without one it slows down. In fact No. A body moves at constant velocity when the resultant force on it is zero (Newton's first law). A resultant force changes the velocity.
Students often think The work done by the applied (or driving) force alone equals the change in the body's kinetic energy, so if the kinetic energy is unchanged that force did no work. In fact Only if it is the only force doing work. The change in kinetic energy equals the work done by the RESULTANT force, which is the sum of the work done by all the forces.
Mechanical energy
Mechanical energy
The sum of the kinetic energy, gravitational potential energy and elastic potential energy of a system. It does not include internal (thermal) energy, chemical energy or other stores. SI unit: joule (J).
Students often think Mechanical energy is just kinetic energy plus gravitational potential energy. In fact Yes. Mechanical energy is the sum of kinetic energy, gravitational potential energy and elastic potential energy.
Students often think Thermal energy is the kinetic energy of particles, so it counts as part of a system's mechanical energy. In fact No. Mechanical energy is kinetic, gravitational potential and elastic potential energy of the bodies as a whole. Internal energy, the random kinetic and potential energy of the particles, is separate.
Non-conservative force
Non-conservative force
A force, such as friction, air resistance (drag) or a push supplied by a person or motor, whose work changes the total mechanical energy of a system. The change in the total mechanical energy of a system equals the work done on it by the non-conservative forces: ΔE_k + ΔE_p + ΔE_H = W_nc. Friction and air resistance do negative work, so mechanical energy decreases and the energy appears as internal energy of the body and its surroundings. Gravity and the elastic force of a spring are conservative forces: their work is accounted for by changes in potential energy.
Conservation of mechanical energy
In the absence of frictional and resistive forces (and of any other non-conservative force doing work), the total mechanical energy of a system is constant. Energy is then only transformed between kinetic, gravitational potential and elastic potential energy, so, for example, a body released from rest reaches a speed v = √(2gΔh) after falling through Δh, whatever its mass and whatever frictionless path it follows.
Students often think Energy is always conserved, so mechanical energy is always conserved too, even with friction. In fact No. Mechanical energy is conserved only when no frictional or resistive (non-conservative) forces do work. When they do, the change in mechanical energy equals the work they do, which is negative.
Students often think The work done by friction and air resistance equals the change in the body's kinetic energy. In fact No. The work done by the resultant force equals the change in kinetic energy; the work done by the non-conservative forces equals the change in the total mechanical energy.
Kinetic energy (E_k)
Kinetic energy (E_k)
The energy a body has because of its translational motion: E_k = ½mv², where m is the mass and v the speed. Because momentum p = mv, E_k can also be written E_k = p²/2m. E_k is a scalar, is never negative, and is proportional to the square of the speed, so doubling the speed quadruples the kinetic energy. SI unit: joule (J).
Gravitational potential energy change (ΔE_p)
The change in gravitational potential energy of a body of mass m when its height changes by Δh close to the surface of the Earth, where the gravitational field strength g is uniform: ΔE_p = mgΔh. It depends only on the vertical change in height, not on the path taken or on the choice of zero level. SI unit: joule (J).
Elastic potential energy (E_H)
The energy stored in a body that has been elastically deformed. For a spring obeying Hooke's law with spring constant k, stretched or compressed by Δx, E_H = ½k(Δx)². The ½ arises because the force rises steadily from zero to kΔx during the deformation, so the work done is the average force ½kΔx multiplied by Δx (the area under the force–extension graph). SI unit: joule (J).
Spring constant (k)
The force per unit extension (or compression) of a spring that obeys Hooke's law, F_H = −kΔx; a stiffer spring has a larger k. SI unit: N m⁻¹.
Students often think Kinetic energy is proportional to speed, so doubling the speed doubles the kinetic energy (the square is dropped). In fact No. E_k = ½mv², so kinetic energy is proportional to the square of the speed: doubling the speed quadruples the kinetic energy.
Students often think Kinetic energy is mv² (the ½ is dropped or seen as unimportant). In fact No. E_k = ½mv². The ½ arises because the speed rises steadily from zero while the resultant force does work over the distance.
Power (P)
Power (P)
The rate at which work is done, or the rate at which energy is transferred: P = ΔW/Δt. For a constant force F acting on a body moving at speed v in the direction of the force, P = Fv. Power is not the same as energy: two machines doing the same work have different powers if they take different times. SI unit: watt (W), where 1 W = 1 J s⁻¹.
Students often think Power and energy mean the same thing, so a more powerful machine transfers more energy, or a fuel that stores more energy releases it faster. In fact No. Energy (J) is an amount; power (W) is the rate of transferring it, P = ΔW/Δt. A more powerful machine transfers energy faster, not necessarily more of it.
Students often think A more powerful machine is one that exerts a larger force, so power can be found from the force alone. In fact No. Power is the rate of doing work, P = Fv; a large force on something that moves slowly may involve little power.
Efficiency (η)
Efficiency (η)
The fraction of the total energy (or power) input to a device that is transferred usefully: η = useful work out / total work in = useful power out / total power in. It has no unit, may be expressed as a percentage, and cannot exceed 1 (100%), since the useful output cannot be greater than the input.
Students often think Efficiency compares useful energy with wasted energy, so it is useful output divided by wasted output. In fact No. Efficiency is the useful output divided by the TOTAL input: η = useful out / total in.
Students often think Efficiency is a measure of the energy lost, so the fraction wasted is the efficiency (and the useful fraction is 1 − η). In fact No. Efficiency is the fraction transferred usefully. The fraction wasted is 1 − η.
Energy density of a fuel
Energy density of a fuel
A measure of how much energy a fuel source releases per unit quantity of fuel. It may be quoted per unit volume (J m⁻³) or per unit mass (J kg⁻¹, also called specific energy), and the two can rank fuels very differently: hydrogen gas releases more energy per kilogram than petrol but far less per cubic metre. The unit given always shows which is meant. Fuel sources differ enormously: complete fission of uranium-235 releases about 8 × 10¹³ J kg⁻¹, over a million times the roughly 3 × 10⁷ J kg⁻¹ released by burning coal.
Students often think A fuel's energy density is a single fixed ranking, so the fuel that releases more energy per kilogram also releases more per litre. In fact Not necessarily. Energy per unit volume equals energy per unit mass multiplied by the fuel's density, so a low-density fuel such as hydrogen gas can rank first per kilogram and last per cubic metre.
Students often think A denser, heavier-feeling fuel contains more energy in each kilogram. In fact No. Energy released per kilogram depends on the chemical (or nuclear) reactions of the fuel, not on its density; hydrogen, the least dense fuel, releases the most energy per kilogram of any chemical fuel.
Diagnostic a bearings check, not a test
10 questions, one per part of the topic where we can. Answer them, then see which statements you own and which to read.
1 Which statement expresses the principle of the conservation of energy?
Answer and reasoning
Energy is conserved until it is used up in doing work, after which it no longer exists. — A student who thinks energy is consumed picks this. Doing work transfers energy; it does not destroy it. After the work is done every joule is still present in some store, often as internal energy of the surroundings.
Each object in a system keeps the same amount of energy throughout any process. — A student who applies conservation to single objects picks this. Energy moves between objects all the time; only the TOTAL energy of an isolated system is constant.
Energy cannot be created or destroyed, so the total energy of an isolated system stays constant. — This is the principle: energy is only transferred between stores and places. Applied to an isolated system, the total energy is the same before and after any process.
The kinetic plus potential energy of any system stays constant, whatever forces act on it. — A student who merges total energy with mechanical energy picks this. Kinetic plus potential energy is conserved only when no friction or other non-conservative force does work; with friction, mechanical energy decreases and internal energy increases.
2 A weightlifter holds a barbell stationary above her head for 4.0 s. How much work does she do on the barbell during these 4.0 s?
Answer and reasoning
A large amount, because holding the heavy barbell up takes a great deal of effort. — A student who equates work with effort picks this. Effort and tiredness come from energy transfers inside the body. On the barbell, the displacement is zero, so the work done on it is zero.
Her upward force multiplied by the 4.0 s for which she holds it up. — A student who thinks work is force multiplied by time picks this. Force × time is impulse, not work. Work is force multiplied by displacement along the force, and the displacement here is zero.
An amount equal to the gravitational potential energy of the barbell. — A student who thinks work is energy a body possesses picks this. The barbell's gravitational potential energy was gained while it was being lifted; during the 4.0 s it is held still, no further energy is transferred to it, so no work is done.
None, because the barbell does not move, so no energy is transferred to it. — Work is done on a body only when the force moves its point of application: W = Fs cos θ with s = 0 gives W = 0. The barbell's energy does not change while it is held still. Her muscles do transfer chemical energy to internal energy in her body, which is why she tires, but none of this goes to the barbell.
3 Which statement is true of every correctly drawn Sankey diagram?
Answer and reasoning
The output arrows are narrower in total than the input, as energy is used up. — A student who thinks devices use up energy picks this. No energy is destroyed; all of the input appears in the outputs, including energy dissipated to the surroundings, so the output widths add up to exactly the input width.
The longest arrow on the diagram represents the largest transfer of energy. — A student who reads a Sankey diagram like a bar chart picks this. Arrow lengths carry no meaning; the amount of energy is shown by the WIDTH of each arrow.
The widths of the output arrows add up to the width of the input arrow. — Width represents energy (or power) to scale, and energy is conserved, so the total of the outputs, useful and wasted, equals the input. This is what a Sankey diagram shows at a glance.
The arrow for the useful output is wider than any arrow for wasted energy. — A student who assumes devices transfer most of their energy usefully picks this. Many devices do not: in a filament lamp the arrow for thermal energy is far wider than the arrow for light.
4 A child pulls a sledge 25 m along level snow using a rope held at 30° above the horizontal. The tension in the rope is a constant 80 N. How much work does the tension do on the sledge?
Answer and reasoning
1.7 × 10³ J — W = Fs cos θ = 80 × 25 × cos 30° = 1732 J ≈ 1.7 × 10³ J. Only the horizontal component of the tension, 80 cos 30° = 69 N, lies along the displacement.
2.0 × 10³ J — A student who multiplies force by distance without the angle gets 80 × 25 = 2000 J. Only the component of the tension along the displacement does work, so the cos 30° factor is needed.
1.0 × 10³ J — A student who uses sin 30° gets 80 × 25 × 0.50 = 1000 J. That is the vertical component of the tension multiplied by the displacement; the component along the (horizontal) displacement is F cos θ.
3.1 × 10² J — A student whose calculator is in radian mode gets cos 30 = 0.154, and 80 × 25 × 0.154 = 309 J. The angle is in degrees; cos 30° = 0.866.
Working W = Fs cos θ = (80 N)(25 m)(cos 30°) = 2000 × 0.866 = 1732 J ≈ 1.7 × 10³ J.
5 A car of mass 1200 kg starts from rest on a level road. The driving force on it is 4000 N and the total resistive force is 1000 N; both are constant. What is the speed of the car after it has travelled 50 m?
Answer and reasoning
15.8 m s⁻¹ — Resultant force = 4000 − 1000 = 3000 N. Work done by the resultant force = 3000 × 50 = 1.5 × 10⁵ J = ΔE_k = ½ × 1200 × v², so v = √(2 × 1.5 × 10⁵ / 1200) = √250 = 15.8 m s⁻¹.
18.3 m s⁻¹ — A student who equates the work done by the driving force alone with the gain in kinetic energy uses 4000 × 50 = 2.0 × 10⁵ J and gets v = 18.3 m s⁻¹. The resistive force does −5.0 × 10⁴ J of work; only the work done by the resultant force equals ΔE_k.
20.4 m s⁻¹ — A student who adds the magnitudes of the forces uses 5000 N and gets v = 20.4 m s⁻¹. The forces act in opposite directions, so the resultant is 4000 − 1000 = 3000 N.
11.2 m s⁻¹ — A student who uses E_k = mv² without the ½ gets v = √(1.5 × 10⁵ / 1200) = 11.2 m s⁻¹. Kinetic energy is ½mv², so v² = 2 × 1.5 × 10⁵ / 1200 = 250.
Working F_net = 4000 N − 1000 N = 3000 N. W_net = F_net s = 3000 N × 50 m = 1.5 × 10⁵ J = ΔE_k = ½mv². v = √(2 × 1.5 × 10⁵ J / 1200 kg) = √(250 m² s⁻²) = 15.8 m s⁻¹.
6 Which quantities are added together to give the mechanical energy of a system?
Answer and reasoning
Its kinetic energy and gravitational potential energy, and no other store. — A student who learned mechanical energy from falling-ball and pendulum examples picks this. Elastic potential energy, as in a spring or bungee cord, is also part of mechanical energy.
Its kinetic energy, gravitational potential energy and elastic potential energy. — Mechanical energy is defined as the sum of kinetic energy, gravitational potential energy and elastic potential energy. Internal, chemical and other stores are not part of it.
Its kinetic energy, gravitational potential energy and internal thermal energy. — A student who counts particle kinetic energy as mechanical energy picks this. Internal energy is separate; when friction acts, mechanical energy decreases and internal energy increases by the same amount.
Its kinetic energy, potential energies and the work that has been done on it. — A student who treats work as a stored form of energy picks this. Work done is a transfer of energy into the stores; adding it to the kinetic and potential energies would count the same energy twice.
7 A skier of mass 60 kg starts from rest at the top of a slope and descends through a vertical height of 20 m. At the bottom her speed is 16 m s⁻¹. Take g = 9.8 m s⁻². What is the total work done on the skier by friction and air resistance during the descent?
Answer and reasoning
+4.1 × 10³ J — A student who believes work cannot be negative drops the sign. The skier ends with less mechanical energy than she started with, so friction and air resistance did negative work on her.
+7.7 × 10³ J — A student who equates the work done by friction and air resistance with the change in kinetic energy gets ½ × 60 × 16² = 7680 J. That is the work done by the resultant force (including gravity). The work done by the non-conservative forces equals the change in total mechanical energy, ΔE_k + ΔE_p.
−4.1 × 10³ J — The change in mechanical energy equals the work done by the non-conservative forces. ΔE_k = ½ × 60 × 16² = +7680 J; ΔE_p = −60 × 9.8 × 20 = −11 760 J. W_nc = 7680 − 11 760 = −4080 J ≈ −4.1 × 10³ J. It is negative because friction and air resistance transfer mechanical energy to internal energy.
−1.1 × 10⁴ J — A student who takes kinetic energy as proportional to speed writes ½ × 60 × 16 = 480 J and gets 480 − 11 760 ≈ −1.1 × 10⁴ J. Kinetic energy is ½mv², so ΔE_k = 7680 J.
Working ΔE_k = ½mv² − 0 = ½(60 kg)(16 m s⁻¹)² = 7680 J. ΔE_p = mgΔh = (60 kg)(9.8 m s⁻²)(−20 m) = −11 760 J. W_nc = ΔE_mech = 7680 J − 11 760 J = −4080 J ≈ −4.1 × 10³ J.
8 Trolley P has a mass of 2.0 kg and trolley Q has a mass of 8.0 kg. The two trolleys have momenta of the same magnitude. How do their kinetic energies compare?
Answer and reasoning
P has four times the kinetic energy of Q. — E_k = p²/2m. With p the same, E_k is inversely proportional to mass, so E_k(P)/E_k(Q) = 8.0/2.0 = 4. P moves four times as fast as Q, and speed enters squared.
P and Q have the same kinetic energy. — A student who equates momentum with kinetic energy picks this. Equal momenta do not give equal kinetic energies: E_k = p²/2m, so the lighter trolley has the larger kinetic energy.
Q has four times the kinetic energy that P has. — A student who thinks the heavier body has more kinetic energy picks this. For equal momenta, Q's speed is a quarter of P's; E_k = p²/2m shows the heavier trolley has LESS kinetic energy.
P has sixteen times the kinetic energy of Q. — A student who squares the speed ratio (4 : 1) but ignores the mass ratio gets 16. E_k = ½mv² depends on both: (2.0/8.0) × 4² = 4.
Working E_k = p²/2m, so with equal p: E_k(P)/E_k(Q) = m_Q/m_P = 8.0 kg / 2.0 kg = 4. Check: v_P = 4v_Q, so E_k(P)/E_k(Q) = (2.0 × 16v_Q²)/(8.0 × v_Q²) = 4.
9 Crane X lifts a load through a height of 10 m in 20 s. Crane Y lifts an identical load through the same height in 40 s. Both lift at constant speed. Which statement is correct?
Answer and reasoning
X does twice as much work as Y, because it transfers energy faster. — A student who treats power and energy as the same thing picks this. Transferring energy faster is greater POWER; the total energy transferred, the work done, is the same for both cranes.
Both do equal work, and X develops twice as much power as Y. — Each crane lifts the same weight through the same height, so each does the same work, mgΔh. Power is the rate of doing work, P = ΔW/Δt; X takes half the time, so it develops twice the power.
Y does more work than X, because it applies its force for twice as long. — A student who thinks work depends on how long a force acts picks this. Work depends on force and displacement; both cranes exert the same force through the same 10 m, so they do the same work.
Both develop the same power, as they exert equal forces on their loads. — A student who equates power with force picks this. Power is P = Fv; the forces are equal, but X lifts its load twice as fast, so X's power is twice Y's.
10 A Sankey diagram for a car engine has an input arrow representing 200 kJ of chemical energy from the fuel. It divides into four output arrows: 60 kJ of useful mechanical work, 100 kJ of thermal energy carried away by the exhaust gases, 30 kJ of thermal energy removed by the cooling system, and 10 kJ of sound. What is the efficiency of the engine?
Answer and reasoning
43% — A student who divides the useful output by the wasted output gets 60 / (100 + 30 + 10) = 60 / 140 = 43%. Efficiency is the useful output divided by the total input, 200 kJ.
70% — A student who thinks efficiency measures the energy wasted gets 140 / 200 = 70%. That is the fraction dissipated; the efficiency is the useful fraction, 60 / 200.
60% — A student who reads the useful arrow's value directly as a percentage gives 60%. That shortcut works only when the input is 100 units; here it is 200 kJ, so 60 kJ is 30% of the input.
30% — η = useful work out / total work in = 60 kJ / 200 kJ = 0.30 = 30%.
Working η = useful work out / total work in = 60 kJ / 200 kJ = 0.30 = 30%. (Check: 60 + 100 + 30 + 10 = 200 kJ, so the outputs equal the input.)
Read the ones marked not yet in Learn, then Verify.
Verify confirm before you go
14 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 A ball is dropped from rest from a height of 2.0 m onto a hard floor and rebounds to a height of 1.5 m. Air resistance is negligible. Which statement correctly accounts for the energy?
Answer and reasoning
The total is unchanged; a quarter of the ball's mechanical energy became internal energy and sound. — At the top of each bounce the ball's mechanical energy is all gravitational potential energy, proportional to height, so it falls from mgh at 2.0 m to mgh at 1.5 m: a loss of 0.5/2.0, one quarter. That energy was transferred during the impact to the internal energy of the ball and floor and to sound. The total energy is conserved.
Energy was not conserved in the bounce, because a quarter of the ball's energy was used up and destroyed. — A student who thinks energy can be used up picks this. The quarter that is missing from the ball's mechanical energy still exists: it was transferred to the internal energy of the ball and floor, and to sound, which is eventually absorbed by the air.
Mechanical energy is conserved, so the missing quarter is still stored as elastic potential energy in the ball. — A student who believes mechanical energy is conserved even when non-conservative forces act picks this. At the top of the rebound the ball is no longer compressed, so it holds no elastic potential energy; during the impact, internal friction in the ball and floor dissipated part of its mechanical energy.
The missing quarter of the ball's energy was changed into the force that the floor exerted on the ball. — A student who treats energy and force as the same kind of thing picks this. The floor's force on the ball is an interaction measured in newtons; it cannot contain energy. The missing energy is in the internal energy of the ball and floor and in sound.
2 Work done and energy are both measured in joules. Which statement correctly defines one joule of work done?
Answer and reasoning
The work done when a force of 1 N acts on a body for a time of 1 s, whether or not the body moves at all. — A student who thinks work depends on how long a force acts picks this. A force of 1 N for 1 s gives an impulse of 1 N s, which is a change of momentum, not a measure of the energy transferred. If the body does not move, no work is done at all.
The work done when a force of 1 N moves its point of application 1 m in the direction of the force. — W = Fs cos θ = 1 N × 1 m × cos 0° = 1 N m = 1 J. Work and energy share the joule because the work done by a force IS the energy it transfers.
The work done when a force of 1 N acts on a body that moves 1 m in any direction relative to that force. — A student who ignores the direction of the force picks this. If the body moves 1 m perpendicular to the 1 N force, the force does no work; only the component along the displacement counts (W = Fs cos θ).
The energy that a force of 1 N contains, whatever body it acts on or however it moves. — A student who treats force and energy as the same thing picks this. A force does not contain energy; it transfers energy only when its point of application moves, and 1 J is transferred by 1 N acting through 1 m.
Working W = Fs cos θ = (1 N)(1 m)(cos 0°) = 1 N m = 1 J.
3 A student draws a Sankey diagram for an electric motor. The input arrow is 50 mm wide and represents 100 J of electrical energy. Only two output arrows are drawn: one 20 mm wide for work done lifting a load, and one 25 mm wide for thermal energy transferred to the surroundings. Which statement about the diagram is correct?
Answer and reasoning
An arrow is missing, but it only represents 5 J, as 1 mm of width stands for 1 J. — A student who reads widths straight off as joules picks this. The scale here is 100 J ÷ 50 mm = 2 J per mm, so the missing 5 mm width represents 10 J, not 5 J.
No arrow is missing; the other 10 J was used up inside the motor and is gone. — A student who thinks energy is used up picks this. The motor cannot destroy energy; the 10 J went somewhere, such as sound or further heating, and a correct Sankey diagram must show it so that the outputs equal the input.
The lifting arrow should be widest, as it shows the motor's useful output. — A student who assumes a device's useful output must dominate picks this. The width is set by the actual energy transferred. A motor that lifts 40 J of every 100 J supplied is realistic, and its useful arrow need not be the widest.
An arrow 5 mm wide is missing; it represents 10 J, since the outputs must total 100 J. — The scale is 100 J ÷ 50 mm = 2 J per mm. The two arrows total 45 mm, which is 90 J, so 10 J (5 mm) is unaccounted for. By conservation of energy the output widths must add up to the input width, so an arrow for this energy, for example sound, is missing.
Working Scale = 100 J / 50 mm = 2 J mm⁻¹. Outputs drawn = 20 mm + 25 mm = 45 mm = 90 J. Missing = 50 mm − 45 mm = 5 mm = 5 mm × 2 J mm⁻¹ = 10 J.
4 A stone tied to a string is whirled in a horizontal circle of radius 0.60 m at constant speed on a smooth, horizontal table. The tension in the string is 12 N. How much work does the tension do on the stone during one complete revolution?
Answer and reasoning
About 45 J, which is the tension multiplied by the full circumference of the circular path. — A student who multiplies force by distance travelled whatever the direction picks this (12 N × 2π × 0.60 m ≈ 45 J). The tension has no component along the stone's direction of motion, so it does no work.
Zero, because the tension is perpendicular to the stone's displacement at every instant. — The tension points to the centre of the circle, while the stone's displacement at each instant is along the tangent, so θ = 90° and W = Fs cos 90° = 0 throughout. This is why the stone's speed, and so its kinetic energy, stays constant.
A positive amount, because the tension is the force that keeps the stone moving. — A student who thinks a force is needed to keep a body moving picks this. The stone would keep moving without the tension, in a straight line; the tension only changes its direction, and a force perpendicular to the motion transfers no energy.
Zero, because the tension is balanced by an equal outward centrifugal force on the stone. — A student who believes a real outward centrifugal force balances the tension picks this: no resultant force, so no work. The reasoning is wrong. In an inertial frame there is no outward force; the resultant force on the stone is the inward tension. It does no work only because it is perpendicular to the stone's displacement at every instant, so W = Fs cos 90° = 0.
5 A trolley moves 4.0 m in a straight line. One of the forces acting on it is a constant 20 N force directed at 120° to the trolley's displacement. How much work does this force do on the trolley?
Answer and reasoning
+40 J — A student who believes work cannot be negative drops the sign. cos 120° = −0.50, so the work done is −40 J: the force takes 40 J of energy from the trolley.
+80 J — A student who multiplies force by distance without the angle gets 20 × 4.0 = 80 J. Only the component along the displacement does work, and here that component points backwards.
+69 J — A student who uses sin 120° = 0.866 gets 20 × 4.0 × 0.866 = 69 J. The work uses the cosine of the angle between the force and the displacement, and cos 120° is negative.
−40 J — W = Fs cos θ = 20 × 4.0 × cos 120° = 80 × (−0.50) = −40 J. The force has a component opposite to the displacement, so it transfers energy away from the trolley.
Working W = Fs cos θ = (20 N)(4.0 m)(cos 120°) = 80 × (−0.50) = −40 J.
6 A student pulls a crate at constant velocity for 5.0 m across a rough, level floor, using a horizontal force of 60 N. Which statement about the work done on the crate is correct?
Answer and reasoning
The pull does 300 J of work on the crate, so the crate's kinetic energy increases by 300 J. — A student who equates the work done by one force with the change in kinetic energy picks this. The pull does 300 J of work, but friction does −300 J; the change in kinetic energy equals the work done by the resultant force, which is zero.
The pull must be greater than friction, so the resultant force does positive work on it. — A student who thinks motion needs a resultant force picks this. The crate moves at constant velocity, so by Newton's first law the resultant force is zero; if the pull exceeded friction the crate would speed up.
The resultant force does no work, so the crate's kinetic energy does not change. — At constant velocity the resultant force is zero: the 60 N pull is balanced by 60 N of friction. The pull does +300 J of work and friction does −300 J, so the work done by the resultant force is zero and the kinetic energy is unchanged. The 300 J ends up as internal energy of the crate and floor.
Friction does no work, because a force opposing the motion cannot do work. — A student who believes work cannot be negative picks this. Friction acts at 180° to the displacement, so it does W = 60 × 5.0 × cos 180° = −300 J of work, transferring energy from the crate to internal energy.
7 Blocks of mass 2.0 kg and 4.0 kg are released from rest at the same height. The 2.0 kg block slides down a steep ramp and the 4.0 kg block down a shallow ramp; the bottoms of the two ramps are at the same level. Both ramps are frictionless and air resistance is negligible. How do their speeds at the bottom compare?
Answer and reasoning
The 4.0 kg block is faster, because its weight is twice as large. — A student who thinks heavier objects fall faster picks this. The 4.0 kg block has twice the weight but also twice the mass to accelerate; its kinetic energy per kilogram, gΔh, is the same as the lighter block's.
The lighter block is faster, because its steeper ramp gives it more acceleration. — A student who equates larger acceleration with larger final speed picks this. The steep ramp does give more acceleration, but over a shorter distance; the speed at the bottom depends only on the height dropped.
The 4.0 kg block is faster, because it accelerates for longer on its ramp. — A student who thinks a longer time accelerating means a greater speed picks this. On the shallow ramp the acceleration is smaller, which exactly offsets the longer time; the final speed is √(2gΔh) on both ramps.
They are equal, as each block loses the same potential energy per kilogram. — With no frictional or resistive forces, mechanical energy is conserved: mgΔh = ½mv², so v = √(2gΔh). Mass cancels and the path shape does not appear, so both blocks reach the same speed.
8 A student of mass 55 kg climbs a staircase from a platform 4.0 m above the ground to a landing 10.0 m above the ground. The staircase is 12 m long, measured along its slope. Take g = 9.8 m s⁻². What is the increase in her gravitational potential energy?
Answer and reasoning
5.4 × 10³ J — A student who uses the final height above the ground gets 55 × 9.8 × 10.0 = 5390 J. The change in potential energy depends on the change in height, 6.0 m, not the height above the ground.
3.2 × 10³ J — ΔE_p = mgΔh with Δh = 10.0 − 4.0 = 6.0 m: ΔE_p = 55 × 9.8 × 6.0 = 3234 J ≈ 3.2 × 10³ J. The length of the staircase does not matter; only the vertical change in height does.
6.5 × 10³ J — A student who uses the distance along the staircase gets 55 × 9.8 × 12 = 6468 J. Her weight acts vertically, so only the vertical rise of 6.0 m counts.
3.3 × 10² J — A student who uses mass in place of weight gets 55 × 6.0 = 330 J. ΔE_p = mgΔh: the mass must be multiplied by g = 9.8 m s⁻² to give the weight in newtons.
Working Δh = 10.0 m − 4.0 m = 6.0 m. ΔE_p = mgΔh = (55 kg)(9.8 m s⁻²)(6.0 m) = 3234 J ≈ 3.2 × 10³ J.
9 A spring of spring constant 180 N m⁻¹ is compressed by 0.10 m and then released to launch a trolley of mass 0.50 kg along a smooth horizontal track. All the elastic potential energy is transferred to the trolley's kinetic energy. What is the speed of the trolley?
Answer and reasoning
6.0 m s⁻¹ — A student who treats elastic energy as proportional to the compression writes ½ × 180 × 0.10 = 9.0 J and gets v = √(2 × 9.0 / 0.50) = 6.0 m s⁻¹. E_H = ½k(Δx)², so Δx must be squared.
2.7 m s⁻¹ — A student who multiplies the final force by the compression writes 18 N × 0.10 m = 1.8 J and gets v = √(2 × 1.8 / 0.50) = 2.7 m s⁻¹. The spring force rises from zero to 18 N, so the energy stored is half of this: 0.90 J.
1.9 m s⁻¹ — E_H = ½k(Δx)² = ½ × 180 × 0.10² = 0.90 J = ½mv², so v = √(2 × 0.90 / 0.50) = √3.6 = 1.9 m s⁻¹.
1.3 m s⁻¹ — A student who uses E_k = mv² without the ½ gets v = √(0.90 / 0.50) = 1.3 m s⁻¹. With ½mv² = 0.90 J, v² = 3.6 m² s⁻².
Working E_H = ½k(Δx)² = ½(180 N m⁻¹)(0.10 m)² = 0.90 J. ½mv² = 0.90 J, so v = √(2 × 0.90 J / 0.50 kg) = √(3.6 m² s⁻²) = 1.9 m s⁻¹.
10 An electric hoist raises a load of weight 1600 N vertically at a constant speed of 0.45 m s⁻¹. What is the power developed by the hoist's upward force on the load?
Answer and reasoning
7.2 × 10² W — At constant speed the upward force equals the weight, 1600 N. P = Fv = 1600 × 0.45 = 720 W = 7.2 × 10² W.
3.6 × 10³ W — A student who divides force by speed gets 1600 / 0.45 = 3556 W. Power is force multiplied by speed, P = Fv, because P = FΔs/Δt.
7.1 × 10³ W — A student who treats the 1600 N weight as a mass and multiplies by g gets 1600 × 9.8 × 0.45 = 7056 W. The load's weight is already a force in newtons.
1.6 × 10³ W — A student who equates power with force gives 1600 W, the size of the force. Power also depends on how fast the force moves its point of application: P = Fv.
Working At constant speed, F = weight = 1600 N. P = Fv = (1600 N)(0.45 m s⁻¹) = 720 W = 7.2 × 10² W.
11 An electric motor with an efficiency of 0.60 takes an input power of 1.5 kW. It lifts a load of mass 90 kg vertically at constant speed. Take g = 9.8 m s⁻². What is the speed of the load?
Answer and reasoning
1.02 m s⁻¹ — Useful power out = η × total power in = 0.60 × 1500 = 900 W. At constant speed P = Fv with F = mg = 90 × 9.8 = 882 N, so v = 900 / 882 = 1.02 m s⁻¹.
1.70 m s⁻¹ — A student who assumes all the input power is useful gets v = 1500 / 882 = 1.70 m s⁻¹. Energy is conserved in total, but 40% of the input is dissipated, so only 900 W lifts the load.
2.83 m s⁻¹ — A student who divides by the efficiency uses 1500 / 0.60 = 2500 W and gets v = 2500 / 882 = 2.83 m s⁻¹. The useful output must be smaller than the input: useful power = 0.60 × 1500 W.
10.0 m s⁻¹ — A student who uses the mass, 90 kg, in place of the weight gets v = 900 / 90 = 10.0 m s⁻¹. The force needed is the weight, mg = 882 N.
Working Useful power out = ηP_in = 0.60 × 1500 W = 900 W. F = mg = (90 kg)(9.8 m s⁻²) = 882 N. v = P/F = 900 W / 882 N = 1.02 m s⁻¹.
12 A coal-fired power station has an electrical power output of 600 MW and an overall efficiency of 0.40. The energy density of the coal it burns is 30 MJ kg⁻¹, that is, each kilogram burned releases 30 MJ. What mass of coal must the station burn each second?
Answer and reasoning
20.0 kg s⁻¹ — A student who assumes all the energy from the coal becomes electrical energy gets 600 / 30 = 20.0 kg s⁻¹. Only 40% of the energy released becomes electrical, so the station needs 1500 MW of input.
8.00 kg s⁻¹ — A student who multiplies the output by the efficiency uses 0.40 × 600 = 240 MW and gets 240 / 30 = 8.00 kg s⁻¹. The input must be LARGER than the 600 MW output: input = 600 / 0.40 = 1500 MW.
50.0 kg s⁻¹ — Total power in = useful power out / η = 600 MW / 0.40 = 1500 MW = 1500 MJ s⁻¹. Mass per second = 1500 MJ s⁻¹ ÷ 30 MJ kg⁻¹ = 50.0 kg s⁻¹.
33.3 kg s⁻¹ — A student who reads the efficiency as the fraction wasted takes the useful fraction as 0.60 and gets 600 / (0.60 × 30) = 33.3 kg s⁻¹. An efficiency of 0.40 means 40% of the input is useful.
Working Input power = output / η = 600 MW / 0.40 = 1500 MW = 1.5 × 10⁹ J s⁻¹. Mass rate = (1.5 × 10⁹ J s⁻¹) / (30 × 10⁶ J kg⁻¹) = 50.0 kg s⁻¹.
13 Hydrogen gas at atmospheric pressure releases 120 MJ of energy per kilogram and 0.011 MJ per litre when burned. Petrol releases 46 MJ per kilogram and 34 MJ per litre. Which conclusion is supported by these data?
Answer and reasoning
A litre of hydrogen releases about 2.6 times the energy of a litre of petrol. — A student who assumes energy per kilogram and energy per litre rank fuels the same way applies the per-kilogram ratio (120 / 46 ≈ 2.6) to volume. The per-litre data show the opposite: petrol releases about 3000 times more per litre.
A litre of petrol releases about 3000 times as much energy as a litre of hydrogen. — Per litre: 34 MJ ÷ 0.011 MJ ≈ 3.1 × 10³, so about 3000 times as much. Although hydrogen releases about 2.6 times more energy per kilogram, its density as a gas is so low that a litre contains very little mass. This is why hydrogen vehicles must store it highly compressed or liquefied.
A kilogram of petrol releases more energy than a kilogram of hydrogen, as petrol is denser. — A student who thinks denser fuels contain more energy per kilogram picks this. The data show hydrogen releases 120 MJ per kilogram against petrol's 46 MJ; density affects energy per litre, not per kilogram.
Hydrogen releases its energy about 2.6 times faster than petrol does when it is burned. — A student who confuses the amount of energy with the rate of transfer picks this. Energy released per kilogram says how MUCH energy a fuel gives, not how quickly; the rate depends on how fast the fuel is burned.
Working Per litre: 34 MJ / 0.011 MJ = 3.1 × 10³ ≈ 3000. Per kilogram: 120 MJ / 46 MJ = 2.6 (hydrogen higher).
14 A block of mass 0.40 kg hangs from a vertical spring of spring constant 50 N m⁻¹. At one instant the block is moving at 1.5 m s⁻¹, its centre is 0.30 m above the level chosen as the zero of gravitational potential energy, and the spring is stretched by 0.20 m from its natural length. Take g = 9.8 m s⁻². What is the mechanical energy of the block–spring–Earth system at this instant?
Answer and reasoning
2.6 J — Mechanical energy is the sum of all three stores: E_k = ½ × 0.40 × 1.5² = 0.45 J, E_p = 0.40 × 9.8 × 0.30 = 1.18 J and E_H = ½ × 50 × 0.20² = 1.0 J, giving 2.6 J. The internal energy of the block and spring is not part of mechanical energy.
1.6 J — A student who takes mechanical energy to be kinetic plus gravitational potential energy only gets 0.45 + 1.18 = 1.6 J. The stretched spring stores elastic potential energy, ½k(Δx)² = 1.0 J, and this is part of the system's mechanical energy.
3.1 J — A student who uses E_k = mv² instead of ½mv² gets 0.90 J for the kinetic energy and a total of 3.1 J. Kinetic energy is ½mv², so the block's kinetic energy is 0.45 J and the total is 2.6 J.
6.6 J — A student who writes the elastic potential energy as ½kΔx, dropping the square, gets ½ × 50 × 0.20 = 5.0 J for the spring and a total of 6.6 J. The extension must be squared: E_H = ½k(Δx)² = 1.0 J.
Working Mechanical energy = E_k + E_p + E_H.
E_k = ½mv² = ½ × 0.40 kg × (1.5 m s⁻¹)² = 0.45 J.
E_p = mgh = 0.40 kg × 9.8 m s⁻² × 0.30 m = 1.176 J.
E_H = ½k(Δx)² = ½ × 50 N m⁻¹ × (0.20 m)² = 1.0 J.
Total = 0.45 + 1.176 + 1.0 = 2.626 J ≈ 2.6 J (2 s.f.).
That was your twenty minutes. Real practice on A.3 is past-paper questions marked against the mark scheme.
Paper 1A tests it as multiple choice; Paper 1B through data you have not seen before; Paper 2 with short answers and, at HL, extended responses. Look for the command words — outline, explain, compare, evaluate — and give exactly what each asks for.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank · Specialist review in progress. How these pages are made · Free, no account ·
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