Work, energy and power

Topic A: Space, Time and Motion

Lesson A.3 Work, Energy and Power | IB Physics HL

Energy methods often solve a motion problem without calculating every force or every instant of the motion. This lesson explains how forces transfer energy, how mechanical energy changes and how power describes the rate of transfer. Begin by choosing the system and identifying the useful output.

Learning objectives

Calculate work done by a constant force and explain positive, negative and zero work.
Determine work from a force–displacement graph.
Use the work–kinetic-energy theorem.
Calculate kinetic, gravitational potential and elastic potential energy using IB notation.
Apply conservation of mechanical energy and include transfers caused by friction or resistance.
Calculate average power and use force × speed when the force is parallel to motion.
Calculate efficiency and interpret Sankey diagrams.
Use energy-density data to estimate fuel requirements.
Evaluate an energy investigation with a graph and uncertainties.

IB scope: A.3 is shared by SL and HL; there is no additional HL content in this section. Gravitational potential-energy calculations here use constant g close to Earth’s surface.

IB formula reference

$W=Fs\cos\theta$

$E_k=\frac12mv^2=\frac{p^2}{2m}$

$\Delta E_p=mg\Delta h$

$E_H=\frac12k(\Delta x)^2$

$P=\frac{\Delta W}{\Delta t}=Fv$

$\eta=\frac{\text{useful work out}}{\text{total work in}}$

$\eta=\frac{\text{useful power out}}{\text{total power in}}$

These use the notation in your formula image. In the power formula, $\Delta W/\Delta t$ gives average power over an interval, while Fv gives instantaneous power for a force parallel to velocity. For steady conditions the two descriptions agree. Efficiency is a fraction; multiply by 100 to express it as a percentage.

Use $g=9.81\,\mathrm{m\,s^{-2}}$ unless stated otherwise. In the work formula, θ is the angle between force and displacement. In the spring formula, $\Delta x$ is extension or compression measured from the spring’s natural length.

A.3.1 — Work done by a force

Work is an energy transfer caused by a force acting through a displacement. For a constant force acting over a straight displacement:

$W=Fs\cos\theta$

F and s are magnitudes. The cosine supplies the sign of the work. Work is a scalar and its SI unit is the joule:

$1\,\mathrm{J}=1\,\mathrm{N\,m}$

The sign of work
Force relative to displacement	Work	Example
Component in the displacement direction	Positive	A forward pull on a moving trolley
Component opposite displacement	Negative	Sliding friction on a crate moving across a stationary floor
Perpendicular to displacement	Zero	Normal force during motion along a horizontal floor
No displacement at the point of force application	Zero mechanical work by that force	Supporting a stationary object

A person holding a heavy bag may use metabolic energy while the bag remains stationary. That does not mean the supporting force does mechanical work on the bag. In ideal uniform circular motion, the inward force is perpendicular to the instantaneous motion and does no work.

Worked example 1 — Pulling at an angle

A constant 50.0 N force pulls a crate through 4.00 m horizontally. The force is 30.0° above horizontal. Calculate the work done by this pull.

Show worked answer

Given: F = 50.0 N, s = 4.00 m and θ = 30.0°.

Step 1 — Choose the component along the displacement.

$W=Fs\cos\theta$

Step 2 — Substitute.

$W=(50.0)(4.00)\cos30.0^\circ=173\,\mathrm{J}$

Final answer: The pull does +173 J of work.

Check: This is less than 200 J because only the horizontal component contributes. It is the work by the pull, not necessarily the net work by all forces.

A.3.2 — Work from a force–displacement graph

The signed area under a graph of force component along the motion against displacement gives the work done by that force. This is especially useful when the force changes. Areas above the axis are positive and areas below it are negative.

Use rectangles, triangles or trapezia for simple graphs. For curved graphs, estimate area using narrow strips. The graph must show the force component along the displacement, rather than a perpendicular force.

For a linearly increasing force, work equals the triangular area, not the maximum force multiplied by the full displacement.

A.3.3 — Kinetic energy and the work–energy theorem

Kinetic energy is the energy associated with motion. It depends on speed squared and is non-negative in this non-relativistic model.

$E_k=\frac12mv^2$

Using $p=mv$ , the equivalent momentum expression is:

$E_k=\frac{p^2}{2m}$

Here p is momentum magnitude. Equal momentum does not imply equal kinetic energy when masses differ.

The work–kinetic-energy theorem states that the net work on a particle or body treated as a point mass equals its change in kinetic energy.

$W_{\mathrm{net}}=\Delta E_k=E_{k,f}-E_{k,i}$

Show the algebraic derivation for constant force in one dimension

For a constant resultant force parallel to the motion, use $F_{\mathrm{net}}=ma$ and $v^2-u^2=2as$ .

$W_{\mathrm{net}}=F_{\mathrm{net}}s=mas$

$W_{\mathrm{net}}=\frac12m(v^2-u^2)$

$W_{\mathrm{net}}=E_{k,f}-E_{k,i}$

The theorem also applies to changing forces when net work is found correctly. This derivation establishes the constant-force case without calculus.

Worked example 2 — A pull with friction

A 2.00 kg trolley starts from rest. It is pulled horizontally by 10.0 N through 3.00 m while a constant 2.00 N resistance opposes motion. Calculate its final speed.

Show worked answer

Step 1 — Calculate the work by each horizontal force.

$W_{\mathrm{pull}}=(10.0)(3.00)=30.0\,\mathrm{J}$

$W_{\mathrm{resistance}}=-(2.00)(3.00)=-6.00\,\mathrm{J}$

Step 2 — Add the work contributions. Weight and normal force do zero work here.

$W_{\mathrm{net}}=30.0-6.00=24.0\,\mathrm{J}$

Step 3 — Apply the work–energy theorem.

$24.0=\frac12(2.00)v^2-0$

$v=\sqrt{24.0}=4.90\,\mathrm{m\,s^{-1}}$

Final answer: Final speed is 4.90 m/s.

Check: The kinetic-energy increase equals net work, not the pull’s 30.0 J alone.

Worked example 3 — Braking distance from energy

A 1000 kg car travels at 20.0 m/s. A constant 4000 N resultant braking force acts opposite its motion until it stops. Find the braking distance.

Show worked answer

Step 1 — Calculate initial kinetic energy.

$E_{k,i}=\frac12(1000)(20.0)^2=2.00\times10^5\,\mathrm{J}$

Step 2 — Equate negative braking work to the kinetic-energy change.

$-4000s=0-2.00\times10^5$

Step 3 — Solve for distance.

$s=50.0\,\mathrm{m}$

Final answer: Braking distance is 50.0 m.

Check: The braking work is negative, but the distance is positive. This excludes the driver’s reaction distance.

A.3.4 — Gravitational potential energy

Gravitational potential energy belongs to the interacting body–Earth system. Near Earth’s surface, where g is approximately constant:

$\Delta E_p=mg\Delta h$

A rise gives positive $\Delta E_p$ ; a fall gives negative $\Delta E_p$ . The zero level is a choice, while the change between two heights is physically meaningful.

$W_g=-\Delta E_p$

During a fall, gravity does positive work as gravitational potential energy decreases. During a rise, gravity does negative work. The work by a lifting force equals the potential-energy increase only when changes in kinetic energy and other transfers are properly accounted for.

A.3.5 — Elastic potential energy

A spring obeying Hooke’s law stores energy when extended or compressed from its natural length:

$E_H=\frac12k(\Delta x)^2$

The squared extension means compression and extension of the same magnitude store equal energy in the ideal linear spring. The expression applies within the regime where k remains constant.

For slow stretching from natural length, the applied force rises from zero to $k\Delta x$ . The triangular area on the force–extension graph gives:

$W_{\mathrm{stretch}}=\frac12(\Delta x)(k\Delta x)=\frac12k(\Delta x)^2$

Between two extensions, calculate the change in spring energy:

$\Delta E_H=\frac12k\big[(\Delta x_f)^2-(\Delta x_i)^2\big]$

Do not replace this with one-half k times the square of the difference of the two extensions.

Worked example 4 — Energy stored in a spring

A spring with constant $200\,\mathrm{N\,m^{-1}}$ is stretched slowly from natural length to 0.100 m. Calculate the energy stored and the final applied-force magnitude.

Show worked answer

Step 1 — Calculate stored elastic energy.

$E_H=\frac12(200)(0.100)^2=1.00\,\mathrm{J}$

Step 2 — Use Hooke’s law for the final force.

$F_{\mathrm{final}}=k\Delta x=(200)(0.100)=20.0\,\mathrm{N}$

Final answers: Stored energy 1.00 J; final applied force 20.0 N.

Check: Work is not 20.0 × 0.100 = 2.00 J, because the force increases from zero. Its average value during this slow stretch is 10.0 N.

A.3.6 — Conservation of energy and system choice

Energy cannot be created or destroyed. It can be transferred between systems and transformed between forms. A decrease in mechanical energy does not imply a decrease in total energy.

$E_{\mathrm{mech}}=E_k+E_p+E_H$

If no mechanical energy enters or leaves through external work and there is no dissipative transfer, mechanical energy is conserved:

$E_{k,i}+E_{p,i}+E_{H,i}=E_{k,f}+E_{p,f}+E_{H,f}$

For a particle model with gravitational and elastic potential energies included, the work by other applied or non-conservative forces changes mechanical energy:

$W_{\mathrm{other}}=\Delta E_{\mathrm{mech}}$

If gravity is represented through $\Delta E_p$ , do not also add its work as an extra energy input. The same applies to a spring represented through $\Delta E_H$ . Sliding friction and resistance usually reduce mechanical energy and increase internal energy of the interacting bodies and surroundings.

Two useful approaches: Use $W_{\mathrm{net}}=\Delta E_k$ when adding work by all forces on a particle. Use a mechanical-energy balance when gravitational and spring interactions are represented by potential energies. Choose one consistent bookkeeping method.

Worked example 5 — Falling without resistance

A 0.500 kg object is released from rest 5.00 m above the landing level. Neglect air resistance. Find the speed immediately before impact.

Show worked answer

Step 1 — Choose the system and reference level. Include the object and Earth; set gravitational potential energy to zero at the landing level.

Step 2 — Apply mechanical-energy conservation.

$mgh=\frac12mv^2$

Step 3 — Cancel mass and calculate.

$v=\sqrt{2gh}$

$v=\sqrt{2(9.81)(5.00)}=9.90\,\mathrm{m\,s^{-1}}$

Final answer: Impact speed is 9.90 m/s, with downward velocity.

Check: Mass cancels, as expected for ideal free fall. The model describes motion before impact, not the collision with the ground.

Worked example 6 — A spring launching a block

A 0.500 kg block rests against a spring compressed by 0.200 m on a horizontal surface. The spring constant is $400\,\mathrm{N\,m^{-1}}$ . A constant 5.00 N friction force acts while the block moves the 0.200 m to the spring’s natural length. Find its speed at that point. The spring is light and obeys Hooke’s law.

Show worked answer

Step 1 — Calculate the initial spring energy.

$E_{H,i}=\frac12(400)(0.200)^2=8.00\,\mathrm{J}$

Step 2 — Calculate the energy transferred by friction.

$W_f=-(5.00)(0.200)=-1.00\,\mathrm{J}$

Step 3 — Apply the energy balance. At natural length the spring energy is zero and height has not changed.

$8.00-1.00=\frac12(0.500)v^2$

$v=\sqrt{28.0}=5.29\,\mathrm{m\,s^{-1}}$

Final answer: Speed at natural length is 5.29 m/s.

Check: Final kinetic energy is 7.00 J. The remaining 1.00 J has become internal energy through friction.

A.3.7 — Power

Power is the rate of energy transfer or work done. Average power over an interval is:

$P_{\mathrm{avg}}=\frac{\Delta W}{\Delta t}$

$1\,\mathrm{W}=1\,\mathrm{J\,s^{-1}}$

For a force parallel to velocity, instantaneous mechanical power is:

$P=Fv$

More generally, use the component of force along velocity:

$P=Fv\cos\theta$

Use the power of the particular force or device being considered. A car moving at constant velocity can receive positive power from its engine while resistance removes mechanical energy at an equal rate. Its net power and kinetic-energy change are then zero.

When a body is lifted vertically at constant speed with negligible resistance, the lifting force equals mg:

$P_{\mathrm{useful}}=mgv$

Worked example 7 — Lifting power and efficiency

A motor lifts a 200 kg load vertically through 6.00 m in 12.0 s at constant speed. Its efficiency is 80.0%. Neglect other resistance. Find useful work, useful power and input power.

Show worked answer

Step 1 — Useful work raises gravitational potential energy.

$W_{\mathrm{useful}}=mg\Delta h=(200)(9.81)(6.00)=1.18\times10^4\,\mathrm{J}$

Step 2 — Find the useful power.

$P_{\mathrm{useful}}=\frac{11772}{12.0}=981\,\mathrm{W}$

Step 3 — Rearrange efficiency.

$\eta=\frac{P_{\mathrm{useful}}}{P_{\mathrm{in}}}$

$P_{\mathrm{in}}=\frac{981}{0.800}=1.23\times10^3\,\mathrm{W}$

Final answers: Useful work 11.8 kJ; useful power 981 W; input power 1.23 kW.

Check: Input power exceeds useful output because the motor also transfers energy to less useful forms.

A.3.8 — Efficiency and Sankey diagrams

Efficiency compares the useful output with the total input. It depends on the purpose of the device. Thermal energy can be a useful output for a heater and an unwanted transfer for a lifting motor.

$\eta=\frac{E_{\mathrm{useful,out}}}{E_{\mathrm{in}}}=\frac{P_{\mathrm{useful,out}}}{P_{\mathrm{in}}}$

The energy and power ratios are equivalent when evaluated consistently over the same interval or under steady operation. In the usual device model, efficiency lies between zero and one.

A Sankey diagram represents energy or power transfers using arrows whose widths are proportional to the transfer amounts. Input equals the sum of outputs, including less useful transfers and any storage change that the model requires.

A.3.9 — Energy density and fuel requirements

Fuel data may express available energy per unit mass or per unit volume. Always read the units. Mass-based values are often called specific energy or gravimetric energy density; volume-based values describe energy per unit volume.

$\text{specific energy}=\frac{E}{m_{\mathrm{fuel}}}$

$\text{volumetric energy density}=\frac{E}{V_{\mathrm{fuel}}}$

Typical units are $\mathrm{J\,kg^{-1}}$ and $\mathrm{J\,m^{-3}}$ , respectively. A high energy density does not by itself determine suitability: storage, availability, cost and environmental effects also matter.

Worked example 8 — Fuel mass for an electrical output

A generator delivers a constant 500 kW electrical output for 1.00 h. Its efficiency is 35.0%, and its fuel releases $46.0\,\mathrm{MJ\,kg^{-1}}$ . Estimate the fuel mass required.

Show worked answer

Step 1 — Convert units and find useful output energy.

$E_{\mathrm{out}}=(5.00\times10^5)(3600)=1.80\times10^9\,\mathrm{J}$

Step 2 — Find total fuel-energy input.

$E_{\mathrm{in}}=\frac{E_{\mathrm{out}}}{\eta}=\frac{1.80\times10^9}{0.350}=5.14\times10^9\,\mathrm{J}$

Step 3 — Divide by energy available per kilogram.

$m_{\mathrm{fuel}}=\frac{E_{\mathrm{in}}}{46.0\times10^6}=112\,\mathrm{kg}$

Final answer: Approximately 112 kg of fuel.

Check: Dividing useful output energy directly by the fuel’s specific energy would ignore the generator’s efficiency.

A.3.10 — Common misconceptions

“A large force always does a large amount of work.” Work also depends on displacement and angle.
“Negative work means negative energy.” Negative net work means kinetic energy decreases; kinetic energy remains non-negative.
“Friction destroys energy.” Mechanical energy transfers mainly to internal energy.
“Spring work equals final force × extension.” A changing force requires its graph area.
“All the work done by the pull becomes kinetic energy.” Add work by every relevant force.
“The same work means the same power.” Power also depends on the time taken.
“Efficiency is the useful output divided by wasted output.” Divide useful output by total input.

A clear energy solution

Choose the system and state the start and end conditions.
List the energy stores and transfers.
State the model: constant force, linear spring, negligible resistance or specified efficiency.
Write the symbolic equation before substituting.
Calculate with consistent SI units and retain extra digits.
Box the final answer and check its sign, units and physical meaning.

A.3.11 — IB-style practice

These original questions combine conceptual reasoning and numerical application. Attempt each question before expanding its answer.

Multiple choice

Multiple choice 1 [1 mark]

A constant force acts perpendicular to a straight displacement. What work does it do?

Fs.
Zero.
−Fs.
The work depends only on the mass.

Show answer and explanation

Answer: B. The angle is 90°, so the component along displacement and its work are zero.

Multiple choice 2 [1 mark]

A body’s speed doubles while its mass remains constant. How does its kinetic energy change?

It doubles.
It halves.
It becomes four times as large.
It stays the same.

Show answer and explanation

Answer: C. Kinetic energy is proportional to speed squared.

Multiple choice 3 [1 mark]

Two particles have the same momentum magnitude. One has twice the mass of the other. How does its kinetic energy compare?

Half as large.
Twice as large.
Four times as large.
The same.

Show answer and explanation

Answer: A. For fixed momentum, kinetic energy equals p²/(2m), so doubling mass halves kinetic energy.

Multiple choice 4 [1 mark]

A spring’s extension doubles within its linear regime. What happens to its stored energy?

It doubles.
It stays the same.
It halves.
It becomes four times as large.

Show answer and explanation

Answer: D. Elastic potential energy is proportional to the square of extension.

Multiple choice 5 [1 mark]

A motor does 600 J of useful work in 3.00 s. What is its average useful power?

1800 W.
200 W.
0.005 W.
600 W.

Show answer and explanation

Answer: B. Average power is work divided by time: 600/3.00 = 200 W.

Multiple choice 6 [1 mark]

A device receives 1000 J and produces 300 J of useful output. What is its efficiency?

70%.
3.33%.
30%.
300%.

Show answer and explanation

Answer: C. Efficiency is useful output divided by total input: 300/1000 = 0.300 = 30%.

Extended-response questions

Show equations, substitutions and units. Mark allocations are suggested guides.

Practice 1 — Work at an angle [3 marks]

A 60.0 N pull acts 40.0° above the direction of a 5.00 m horizontal displacement. Calculate its work and explain why it is less than 300 J.

Show worked answer

Use $W=Fs\cos\theta$ . [1]

$W=(60.0)(5.00)\cos40.0^\circ=230\,\mathrm{J}$

Correct positive work. [1] Only the component parallel to the displacement contributes. [1]

Practice 2 — Negative work [4 marks]

A 2.00 kg object has initial speed 6.00 m/s. The net work on it is −20.0 J. Find its final speed and interpret the negative work.

Show worked answer

$E_{k,i}=\frac12(2.00)(6.00)^2=36.0\,\mathrm{J}$

Correct initial energy. [1]

$E_{k,f}=36.0-20.0=16.0\,\mathrm{J}$

Correct energy balance. [1]

$v=\sqrt{\frac{2(16.0)}{2.00}}=4.00\,\mathrm{m\,s^{-1}}$

Correct speed. [1] Negative net work means kinetic energy decreases. [1]

Practice 3 — A changing force [4 marks]

The force component along motion increases linearly from 10.0 N at displacement zero to 30.0 N at 4.00 m. Find its work. If it is the resultant force on a 2.00 kg body starting from rest, find final speed.

Show worked answer

Use trapezium area. [1]

$W=\frac{10.0+30.0}{2}(4.00)=80.0\,\mathrm{J}$

Correct work. [1]

Apply $W_{\mathrm{net}}=\Delta E_k$ . [1]

$v=\sqrt{\frac{2(80.0)}{2.00}}=8.94\,\mathrm{m\,s^{-1}}$

Correct speed. [1]

Practice 4 — Spring-energy change [4 marks]

A spring of constant $300\,\mathrm{N\,m^{-1}}$ is stretched slowly from extension 0.100 m to 0.200 m. Find the increase in stored energy and the work done by the spring during this additional stretch.

Show worked answer

Use the difference of stored energies. [1]

$\Delta E_H=\frac12(300)\big[(0.200)^2-(0.100)^2\big]=4.50\,\mathrm{J}$

Correct energy increase. [1] Spring work equals $-\Delta E_H$ . [1] Therefore its work is −4.50 J. [1]

Practice 5 — Energy with a dissipative transfer [4 marks]

A 2.00 kg body starts from rest and descends 3.00 m in height. During its motion, 10.0 J of mechanical energy is transferred to internal energy by resistance. Find its final speed.

Show worked answer

The decrease in gravitational potential energy is available for kinetic energy and the dissipative transfer. [1]

$\Delta E_p=mg\Delta h=(2.00)(9.81)(-3.00)=-58.86\,\mathrm{J}$

The potential-energy decrease has magnitude 58.86 J. [1]

$E_{k,f}=58.86-10.0=48.86\,\mathrm{J}$

Correct balance. [1]

$v=\sqrt{\frac{2(48.86)}{2.00}}=6.99\,\mathrm{m\,s^{-1}}$

Correct speed. [1]

Practice 6 — Power at constant speed [4 marks]

A car travels at constant speed 15.0 m/s against a total resistance of 800 N on a level road. Find the mechanical power delivered by the driving force and explain why kinetic energy does not increase.

Show worked answer

The driving force balances resistance, so its magnitude is 800 N. [1] Use $P=Fv$ . [1]

$P=(800)(15.0)=1.20\times10^4\,\mathrm{W}=12.0\,\mathrm{kW}$

Correct driving power. [1] Resistance transfers energy out at the same rate, so net work and kinetic-energy change are zero. [1]

Practice 7 — Efficiency and output accounting [4 marks]

A device receives 500 J. Its useful mechanical output is 150 J and its other output is thermal energy. Find efficiency and thermal output. State the arrow-body width ratio for input, useful output and thermal output on a Sankey diagram.

Show worked answer

Use useful output divided by total input. [1]

$\eta=150/500=0.300=30.0\%$

Correct efficiency. [1] Thermal output is 500 − 150 = 350 J. [1] Arrow-body widths should have ratio 500 : 150 : 350, or 10 : 3 : 7. [1]

Practice 8 — Fuel requirement [4 marks]

A machine must provide 2.00 MJ of useful output. Its efficiency is 25.0% and the fuel’s specific energy is $40.0\,\mathrm{MJ\,kg^{-1}}$ . Find input energy and fuel mass.

Show worked answer

Use $E_{\mathrm{in}}=E_{\mathrm{out}}/\eta$ . [1]

$E_{\mathrm{in}}=2.00/0.250=8.00\,\mathrm{MJ}$

Correct input energy. [1] Divide by energy per kilogram. [1]

$m_{\mathrm{fuel}}=8.00/40.0=0.200\,\mathrm{kg}$

Correct fuel mass. [1]

A.3.12 — Lab investigation: work and kinetic energy

Research question

How does the squared speed of a trolley released from rest depend on its vertical drop, and how does this test an energy model?

Release a low-friction trolley from rest at several heights on the same track. Measure the vertical drop of the trolley’s centre of mass and its final speed with a light gate or calibrated motion sensor. Keep the trolley, release method and measurement position fixed. Repeat each trial and catch the trolley safely beyond the detector.

Illustrative mean results
Vertical drop h / m	Final speed v / m s⁻¹
0.100	1.35
0.200	1.91
0.300	2.34
0.400	2.70

For the last speed measurement, take $\Delta v=0.03\,\mathrm{m\,s^{-1}}$ .

Show that the ideal particle model predicts a straight graph of $v^2$ vertically against h horizontally.
Use the first and last points to estimate the gradient and compare it with the ideal prediction.
Estimate the percentage of initial gravitational potential energy appearing as translational kinetic energy using the last row.
Estimate the absolute uncertainty in the last $v^2$ value.
Suggest two reasons for the difference from the ideal prediction and one improvement.

Show investigation answers

(a) For release from rest with negligible loss and negligible wheel rotation:

$mgh=\frac12mv^2$

$v^2=2gh$

The ideal gradient is $2g=19.62\,\mathrm{m\,s^{-2}}$ and the ideal intercept is zero.

(b)

$\text{gradient}=\frac{(2.70)^2-(1.35)^2}{0.400-0.100}=18.225\,\mathrm{m\,s^{-2}}$

This is approximately 18.2 m/s², below the ideal prediction. A best-fit line through all points is preferable in a full analysis.

(c) Compare translational kinetic energy with gravitational potential-energy decrease:

$\frac{E_k}{mgh}=\frac{v^2}{2gh}$

$\frac{(2.70)^2}{2(9.81)(0.400)}=0.929$

Approximately 92.9% appears as translational kinetic energy in the last trial. The rest need not all be dissipated: some may become rotational kinetic energy of the wheels.

(d) Approximate propagation gives:

$\Delta(v^2)\approx2v\Delta v$

$\Delta(v^2)=2(2.70)(0.03)=0.16\,\mathrm{m^2\,s^{-2}}$

(e) Track friction and air resistance dissipate mechanical energy; wheel rotation also takes a share of the available energy. An incorrect vertical-height measurement or an unintended push can bias results. Use low-inertia wheels, measure centre-of-mass drop carefully, calibrate the detector and use a consistent release. Repeats assess random scatter. Rotational energy is developed in A.4.

A.3.13 — Using our strength to serve

Scripture — Colossians 3:23, KJV

“And whatsoever ye do, do it heartily, as to the Lord, and not unto men.”

Creation: Christians see the ordered world and our ability to understand it as gifts from God. Learning how energy is transferred can help us care for resources, design useful solutions and appreciate creation.

The Fall: Our work can become selfish, careless or dishonest. Scripture calls us to recognize sin and turn toward God. In science, honesty means recording what we observe rather than changing results to obtain the answer we hoped for.

Redemption: Jesus Christ died for our sins and rose again. We receive forgiveness through His grace, not through our achievements. You are invited to trust Him as your Saviour and Lord and allow His love to guide your work.

Restoration: Christian hope gives purpose to faithful service. Use your knowledge and energy to help others and care for the world entrusted to us. Read the Bible, spend time in prayer and grow alongside other believers.

Reflection: How can you work wholeheartedly while treating others with kindness? What practical change could help you use time and resources more responsibly?

Prayer: Lord, help me learn with perseverance and work with integrity. Thank You for Your grace through Jesus. Guide me to use my abilities to serve others and honour You. Amen.

Lesson summary and readiness checklist

Work depends on force, displacement and their angle.
Force–displacement area gives work.
Net work equals the change in kinetic energy.
Use the IB energy notation $E_k$ , $E_p$ and $E_H$ .
Mechanical energy is conserved only under the appropriate transfer assumptions.
Resistance transfers mechanical energy mainly to internal energy.
Power is the rate of work or energy transfer.
Efficiency compares useful output with total input.
Sankey arrow widths represent energy or power amounts.
Fuel calculations require consistent energy-density units and efficiency.

Next lesson: A.4 — Rigid Body Mechanics extends motion and energy ideas to rotation.

IB Physics HL Book 2025; Papers 1 & 2 Practice 5 Sections65 Lessons54 Weeks

Curriculum

Work, energy and power

Topic A: Space, Time and Motion

Learning objectives

IB formula reference

A.3.1 — Work done by a force

Worked example 1 — Pulling at an angle

A.3.2 — Work from a force–displacement graph

A.3.3 — Kinetic energy and the work–energy theorem

Worked example 2 — A pull with friction

Worked example 3 — Braking distance from energy

A.3.4 — Gravitational potential energy

A.3.5 — Elastic potential energy

Worked example 4 — Energy stored in a spring

A.3.6 — Conservation of energy and system choice

Worked example 5 — Falling without resistance

Worked example 6 — A spring launching a block

A.3.7 — Power

Worked example 7 — Lifting power and efficiency

A.3.8 — Efficiency and Sankey diagrams

A.3.9 — Energy density and fuel requirements

Worked example 8 — Fuel mass for an electrical output

A.3.10 — Common misconceptions

A clear energy solution

A.3.11 — IB-style practice

Multiple choice

Multiple choice 1 [1 mark]

Multiple choice 2 [1 mark]

Multiple choice 3 [1 mark]

Multiple choice 4 [1 mark]

Multiple choice 5 [1 mark]

Multiple choice 6 [1 mark]

Extended-response questions

Practice 1 — Work at an angle [3 marks]

Practice 2 — Negative work [4 marks]

Practice 3 — A changing force [4 marks]

Practice 4 — Spring-energy change [4 marks]

Practice 5 — Energy with a dissipative transfer [4 marks]

Practice 6 — Power at constant speed [4 marks]

Practice 7 — Efficiency and output accounting [4 marks]

Practice 8 — Fuel requirement [4 marks]

A.3.12 — Lab investigation: work and kinetic energy

Research question

A.3.13 — Using our strength to serve

Scripture — Colossians 3:23, KJV

Lesson summary and readiness checklist

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