Energy stores, work and kinetic energy
Key idea: Energy accounting becomes reliable when you choose a system and name the transfers across its boundary. Work is the mechanical route by which a force transfers energy.
Continue where you stopped
The core idea
H1 Physics 8867 · Lesson 1 of 3
Check your understandingBy the end of this lesson, you should be able to
- Identify energy stores and transfers and apply conservation.
- Calculate work using the displacement component along a force.
- Derive and use Ek = ½mv².
Learn the idea
Big question: How does choosing a system turn a vague energy story into a calculation you can trust?
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
- State the system before describing where energy goes.
- Friction transfers energy to internal stores; it does not destroy energy.
- Net work may be negative, reducing kinetic energy.
Relationships to know
W = Fs cos θWnet = ΔEkEk = ½mv²
Follow 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.
Now try it with support
Practise with support
A 5.0 kg trolley speeds up from 2.0 to 6.0 m s⁻¹. Find the net work done.
Hints
- Use the change in kinetic energy, not the final kinetic energy alone.
- Keep both squared speeds inside the subtraction.
View the guided answer
Wnet = ½(5.0)(6.0² − 2.0²) = 80 J.
Your turn
Practise independently
A 2.0 kg block starts from rest and is pulled 5.0 m by a 14 N horizontal force against 6.0 N friction. Find its final speed.
Check your answer
The resultant force is 14 − 6 = 8.0 N, so net work over 5.0 m is 40 J. From ½(2.0)v² = 40, v = √40 = 6.32 m s⁻¹, or 6.3 m s⁻¹.
Common mistakes and exam guidance
Watch out for
- Adding frictional work as positive when it opposes the displacement.
- Using W = Fs without checking the angle between force and displacement.
In an exam
- A complete energy answer names the initial store, transfer and final stores.
- Check the sign of work against whether the speed should rise or fall.
Put the ideas together
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.
View the marking points and model answer
Marking points
- Calculates ΔEk = 1.536 × 10⁵ J.
- Calculates ΔEp = 2.119 × 10⁵ J.
- Adds the useful store increases.
- Uses engine work minus these increases.
- Obtains about 8.45 × 10⁴ J.
- Identifies internal-energy stores/heating rather than destroyed energy.
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.
Finish from memory
Three-question recap
What does negative work do to kinetic energy?
Check
It reduces kinetic energy if the total net work is negative.
Why is work by a perpendicular force zero?
Check
It has no component along the displacement.
State the work–energy theorem.
Check
Net work on a body equals its change in kinetic energy.
Continue with the next resource in this course.
Course and syllabus information
- Course
- GCE A-Level H1 Physics
- Edition
- GCE A-Level H1 Physics 2027