Energy stores, work and kinetic energy
Key idea: H2 Physics lessons on energy stores, work, kinetic and potential energy, fields, power and efficiency.
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The core idea
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Big question: How does work track energy transferred by a force?
Work is energy transferred when a force acts through a displacement: W = Fs cosθ for a constant force. The work–energy theorem says the net work on an object equals its change in kinetic energy. A force–displacement graph extends this idea: signed area under the graph gives work, even when force changes.
Name stores and transfers precisely
Energy is conserved, but an energy store inside your chosen system can increase or decrease. Begin by naming the system, then identify energy crossing its boundary by mechanical work, electrical work, heating or radiation.
Friction does not destroy energy. It usually transfers energy from a mechanical store into internal-energy stores of the surfaces and surroundings.
Check your understanding: A braking bicycle slows on a level road. Where does its kinetic energy go?
Mainly into internal-energy stores of the brakes, tyres, road and surrounding air through heating and deformation.
Connect net work to a change in speed
For a constant force, W = Fs cos θ uses only the component along the displacement. Work is positive when the force helps the motion and negative when it opposes it.
The net work on a body equals its change in kinetic energy. This follows from Fs = mas and v² − u² = 2as, giving Wnet = ½mv² − ½mu².
Check your understanding: A force is always perpendicular to an object's displacement. What work does that force do?
Zero, because cos 90° = 0. The force can change direction without changing kinetic energy.
Key ideas to keep
- Use the angle between force and displacement, not an angle copied from the diagram.
- Negative work means the force removes kinetic energy from the chosen object.
- Area under an F–x graph is work; its gradient is not force.
See the reasoning
Worked example
Use signed work from several forces
Question: A 4.0 kg trolley starts at 3.0 m s⁻¹. It moves 5.0 m while a 12 N pull acts 30° above the direction of motion and friction is 4.0 N. Find its final speed.
Step 1: Find each mechanical transfer
Why: Only the force component parallel to displacement does work.
Working: Wpull = 12(5.0)cos30° = 52.0 J; Wfriction = −4.0(5.0) = −20.0 J.
Step 2: Find net work
Why: The change in kinetic energy comes from the signed total.
Working: Wnet = 52.0 − 20.0 = 32.0 J.
Step 3: Apply the work–energy theorem
Why: Initial kinetic energy must be retained when the trolley is already moving.
Working: ½(4.0)v² − ½(4.0)(3.0²) = 32.0, so 2v² = 50.0.
Answer: v = 5.0 m s⁻¹. The pull transfers 52 J mechanically; 20 J goes to internal stores and the kinetic store rises by 32 J.
Check: The speed rises because the net work is positive, and the final kinetic energy 50 J equals 18 J + 32 J.
Another worked model
Question
Derive Eₖ = ½mv² for a body accelerated from rest by a constant resultant force.
Check the worked solution
W = Fs and F = ma, so W = mas. From v² = u² + 2as with u = 0, as = v²/2. Therefore W = m(v²/2) = ½mv². The work is the kinetic-energy increase.
Use a hint if needed
Practise with support
Try this
A 20 N force acts 60° to a 5.0 m displacement. Find the work done and the kinetic-energy increase if it is the only transfer.
Hint: Use the force component along displacement.
Check your answer
W = Fs cos 60° = 20(5.0)(0.5) = 50 J, so kinetic energy increases by 50 J.
Now work without the hint
Practise independently
Your turn
A 1200 kg car slows from 20 to 10 m s⁻¹. Find the kinetic-energy change and describe where that energy may be transferred during braking.
Check your answer
ΔEₖ = ½(1200)(10² − 20²) = −1.80 × 10⁵ J. Energy is transferred mainly to thermal stores in brakes, tyres, road and surrounding air.
Avoid these traps
Common mistakes
Common mistake
An inefficient device destroys energy.
What is wrong with this reasoning?
Show better thinking
Energy is conserved. Inefficiency means some input is transferred to less useful stores, commonly thermal stores in the device and surroundings.
Common mistake
Every force does work equal to force times path length.
What is wrong with this reasoning?
Show better thinking
For a constant force, use displacement in the force direction: W = Fs cos θ. A perpendicular force does no work.
Write for the examiner
Exam guidance
Define the system and compare initial and final stores before substituting numbers.
Exam-style practice [6 marks]
A 1200 kg car climbs a hill through a vertical height of 18 m. Its speed rises from 12 m s⁻¹ to 20 m s⁻¹ while its engine does 4.50 × 10⁵ J of work. Find the energy transferred to internal stores. Use g = 9.81 m s⁻².
Plan before you answer
- Choose car and Earth as the system.
- Calculate both the kinetic and gravitational-store increases.
- Use conservation to find the unaccounted transfer.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
ΔEk = ½(1200)(20² − 12²) = 1.536 × 10⁵ J and ΔEp = 1200(9.81)(18) = 2.119 × 10⁵ J. The increase in mechanical stores is 3.655 × 10⁵ J, so 4.50 × 10⁵ − 3.655 × 10⁵ = 8.45 × 10⁴ J is transferred to internal-energy stores of the car and surroundings.
Come back in three days
Check what stayed with you
Recall question 1
What does negative work do to kinetic energy?
Check the answer
It reduces kinetic energy if the total net work is negative.
Recall question 2
Why is work by a perpendicular force zero?
Check the answer
It has no component along the displacement.
Recall question 3
State the work–energy theorem.
Check the answer
Net work on a body equals its change in kinetic energy.
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. For this topic, a field is a region where a mass, charge or current-carrying conductor experiences a force. Keep field force, potential energy and power ideas distinct, and state the system whenever you apply conservation of energy.
- GCE A-Level H2 PhysicsTopic 4(a) / Topic 4(b) / Topic 4(c) / Topic 4(d) / Topic 4(e) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027