Potential energy, power and efficiency
Key idea: H2 Physics lessons on energy stores, work, kinetic and potential energy, fields, power and efficiency.
Continue where you stopped
The core idea
Build the idea
Learn the idea
Big question: How quickly is energy transferred, and how much becomes useful output?
Gravitational, electric and elastic potential energy belong to different interactions. Elastic energy is the area under a force–extension graph. Power is the rate of energy transfer, P = ΔE/Δt, and mechanical power is P = Fv cosθ. Efficiency compares useful output with total input using energy or power over the same interval.
Name the interaction behind each potential store
Gravitational potential energy belongs to interacting masses, electric potential energy to interacting charges and elastic potential energy to a deformed material. The labels are not interchangeable even though energy can be transferred between these stores.
Elastic potential energy equals the area under a force–extension graph. The triangle ½Fx applies only to a straight Hooke-law line from the origin; a curved graph needs its actual area.
Check your understanding: Why can ½Fx fail for a deformed material?
It assumes force rises linearly from zero with extension; a non-linear graph does not have a triangular area.
Keep power separate from energy
Power is the rate of energy transfer, P = ΔE/Δt. For a force acting on a body, instantaneous mechanical power is P = Fv cosθ, using the force component along velocity.
A powerful device transfers energy quickly; it is not necessarily efficient. Always include time when moving between energy and power.
Check your understanding: What power is transferred by a force perpendicular to velocity?
Zero, because its component along velocity is zero.
Use efficiency as a complete energy account
Efficiency is useful output divided by total input, using either energies over the same process or powers over the same interval. The result has no unit and cannot exceed one for a correctly defined system.
Less useful output is not destroyed energy. Identify the internal-energy, sound or other stores receiving it and explain the practical consequence where relevant.
Check your understanding: A motor receives 500 W and delivers 350 W usefully. Find efficiency and less useful power.
Efficiency = 350/500 = 0.70 or 70%; less useful power = 150 W.
Key ideas to keep
- Use the actual graph area when force is not proportional to extension.
- Power measures a rate, not an amount of energy.
- Efficiency has no unit and cannot exceed one for a correctly defined system.
See the reasoning
Worked example
Find elastic energy from a piecewise graph
Question: Force rises linearly from 0 to 20 N over the first 0.10 m, then linearly to 50 N at 0.20 m. Find the stored elastic energy.
Step 1: Use the graph area
Why: Stored energy is accumulated work, not the final force multiplied by extension.
Working: First triangle = ½(0.10)(20) = 1.0 J.
Step 2: Add the second region
Why: The 0.10–0.20 m section is a trapezium.
Working: Second area = ½(20 + 50)(0.10) = 3.5 J.
Step 3: Interpret the total
Why: Every part of the loading path contributes work.
Working: Eelastic = 1.0 + 3.5 = 4.5 J.
Answer: The material stores 4.5 J.
Check: The value is below Fx = 10 J because force was below its final value throughout loading.
Combine mechanical power and efficiency
Question: A motor pulls with 600 N along the motion at 2.5 m s⁻¹ while receiving 2.0 kW. Find useful power and efficiency.
Step 1: Find the useful transfer rate
Why: Force is parallel to velocity.
Working: Puseful = Fv = 600(2.5) = 1500 W.
Step 2: Form the efficiency ratio
Why: Use useful output divided by total input.
Working: η = 1500/2000 = 0.75.
Step 3: Account for the remainder
Why: Conservation requires another output route.
Working: Less useful power = 2000 − 1500 = 500 W, mainly to internal-energy stores.
Answer: Useful power is 1.5 kW and efficiency is 75%.
Check: Useful output is below input, so the efficiency is physically possible.
Another worked model
Question
A nonlinear force–extension graph is a triangle from 0 to 0.10 m and 30 N, then a trapezium to 0.16 m and 42 N. Find stored elastic energy.
Check the worked solution
Energy is total area: ½(0.10)(30) + ½(30 + 42)(0.06) = 1.50 + 2.16 = 3.66 J. Do not use ½Fx for the whole nonlinear graph.
Use a hint if needed
Practise with support
Try this
A 600 N force drives a vehicle at 12 m s⁻¹ in the same direction. Its engine input is 9.0 kW. Find mechanical power and efficiency.
Hint: Calculate Fv before comparing output with input.
Check your answer
Useful power = Fv = 600(12) = 7200 W. Efficiency = 7200/9000 = 0.80 or 80%.
Now work without the hint
Practise independently
Your turn
Distinguish gravitational, electric and elastic potential energy. Then explain why a 65%-efficient device still conserves energy.
Check your answer
They arise from position in gravitational or electric interactions, or deformation of a material. At 65% efficiency, 65% of input reaches the useful output and 35% reaches less useful stores; total energy remains conserved.
Avoid these traps
Common mistakes
Common mistake
Elastic energy is always force multiplied by extension.
What is wrong with this reasoning?
Show better thinking
Elastic energy is the area under the force–extension graph. The triangular result ½Fx applies only to a straight line from the origin.
Common mistake
Efficiency is useful output divided by wasted output.
What is wrong with this reasoning?
Show better thinking
Efficiency is useful output divided by total input, using energy or power consistently. It cannot exceed 1 or 100%.
Write for the examiner
Exam guidance
Write an energy-flow statement before calculating; it makes the useful and total quantities unambiguous.
Exam-style practice [8 marks]
A spring's force rises linearly from zero to 80 N at 0.16 m extension. A motor stretches it in 0.40 s while taking a constant input power of 20 W. Find the spring's energy increase, useful power, efficiency and energy transferred to other stores.
Plan before you answer
- Use area under the force–extension graph.
- Divide useful energy by time.
- Compare useful and input transfers over the same interval.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
Elastic energy = ½Fx = ½(80)(0.16) = 6.4 J. Useful power = 6.4/0.40 = 16 W, so efficiency = 16/20 = 0.80 or 80%. Input energy = 20(0.40) = 8.0 J, leaving 8.0 − 6.4 = 1.6 J transferred mainly to internal-energy stores.
Come back in three days
Check what stayed with you
Recall question 1
What does area under an F–x graph give?
Check the answer
Elastic potential energy stored, or work done in deformation.
Recall question 2
State mechanical power for an angled force.
Check the answer
P = Fv cosθ.
Recall question 3
Define efficiency.
Check the answer
Useful energy or power output divided by total energy or power input.
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(j) / Topic 4(k) / Topic 4(l) / Topic 4(m) / Topic 4(n) · 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