Energy Calculations: KE, GPE and Conservation
Key idea: Solve O Level kinetic and gravitational potential energy questions using conservation of energy, clear working, vertical height and dissipated energy.
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The core idea
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Learning objectives
- Recognise kinetic, potential, nuclear and internal energy stores
- Describe mechanical energy transfer by a force acting over a distance
- Describe electrical energy transfer by an electric current
- Describe energy transfer by heating due to a temperature difference
- Describe energy transfer by electromagnetic and mechanical waves
- Recall and apply Ek = ½mv² in new situations
- Recall and apply Ep = mgh near the Earth's surface in new situations
- State and apply the principle of conservation of energy
- Recall and apply work done = force × distance moved in the force direction
- Recall and apply power = energy transfer / time taken
- Calculate efficiency as useful energy output / total energy input
- Evaluate prescribed electricity-generation resources by efficiency, cost, reliability and environmental impact
1. Definition
Energy calculations using conservation
Energy calculations use the principle of conservation of energy to relate changes in energy stores (especially GPE and KE) and any energy dissipated to the surroundings.
Review Energy (Stores, Transfers & Conservation) before attempting the questions.
2. Key Ideas
- Core equations (O Level):
- kinetic energy: Eₖ = (1/2)mv²
- change in gravitational potential energy near Earth: Δ Eₚ = mgΔ h
- Eₖ and Δ Eₚ are in joules (J); m is mass in kilograms (kg), v is speed in m s⁻¹, g is gravitational field strength in N kg⁻¹, and Δ h is the vertical height change in metres (m). The numerical value of g may also be written in m s⁻².
- Start by identifying the initial and final energy stores.
- Define the system and reference level when they matter. For GPE calculations, only Δ h matters; the zero level may be chosen for convenience.
- For the common KE–GPE model, write: initial GPE + initial KE = final GPE + final KE + energy dissipated.
- State what is neglected. For example, “air resistance is negligible” means there is no dissipated-energy term for the flight.
- Keep units consistent and convert early: g → kg and cm → m.
- Use the value of g given in the question.
3. Detailed Explanations
A quick method for energy questions
- Mark the initial and final states, including the vertical height change.
- List the energy stores at each state and any energy dissipated.
- Write the conservation equation in words before using formulas.
- Substitute SI values, solve, and check the unit and size of the answer.
For example, if an object falls from rest through vertical height h with negligible air resistance, mgh = (1/2)mv² ⇒ v = square root of 2gh
The mass cancels because, under these assumptions, both the GPE decrease and KE increase are proportional to mass.
Case study 1: Oscillating pendulum
Choose the lowest point B as the zero-GPE reference. At the turning point A, the bob is momentarily at rest, so its KE is zero and its GPE is maximum. With negligible resistance, the GPE decrease from A to B equals the KE increase.
For an ideal pendulum with negligible air resistance and negligible friction at the pivot:
- A → B: GPE decrease = KE increase
- B → D: KE decrease = GPE increase
In equation form:
From point A to point B,
E_(p,A) = E_(k,B)
From point B to point D,
E_(k,B) = E_(p,D)
For a real pendulum, air resistance and friction transfer energy from the pendulum’s mechanical stores to internal energy stores of the pendulum and surroundings. Some energy is also transferred away by sound. The next turning point is therefore lower, although total energy is conserved.
Across the real swing from A to the lower turning point D′,
E_(p,A) = E_(p,D') + E_(dissipated, A to D')
Case study 2: Bouncing ball
Consider a ball of mass m dropped from height hᵢₙᵢₜᵢₐₗ. Choose the ground as the zero-GPE reference and neglect air resistance during each flight.
Stage 1: Initially, the ball has:
- GPE is mghᵢₙᵢₜᵢₐₗ.
- KE is zero because the ball starts from rest.
Stage 2: Just before the ball touches the ground, the ball has:
- GPE is approximately zero at the chosen reference level.
- KE is mghᵢₙᵢₜᵢₐₗ because the GPE decrease equals the KE increase.
Stage 3: When the ball hits the ground and rebounds:
- The ball briefly deforms, storing energy elastically before regaining its shape.
- Some energy is transferred to internal energy stores of the ball and ground, and some is carried away by sound waves.
- The KE just after rebound is therefore smaller than the KE just before impact.
Stage 4: The ball rises to a smaller maximum height, hₘₐₓ:
- GPE is mghₘₐₓ and KE is zero.
- Because energy was dissipated during the collision, hₘₐₓ < hᵢₙᵢₜᵢₐₗ.
If a question includes air resistance, add energy dissipated during the flight as well as during the collision.
4. Common Mistakes
- Using g = 10 without checking what the question states.
- Forgetting unit conversions (g → kg and cm → m).
- Forgetting to include an energy dissipated term when friction/air resistance is mentioned.
- Using the distance along a slope as h (for GPE, h is the vertical height change).
- Writing “energy is lost” as if it has been destroyed. Instead, state the store that decreases and where energy is transferred.
5. Exam Tips
- Start with a one-line equation: initial GPE + initial KE = final GPE + final KE + energy dissipated. Remove any zero terms only after writing the full model.
- For an object dropped from rest, with the ground as the zero-GPE reference and negligible air resistance, its energy just before impact is all KE.
- If resistive forces are negligible, decrease in GPE = increase in KE is often the fastest route.
- If the object is at rest (e.g. at maximum height), set Eₖ = 0.
- Always give a final line with a unit (J for energy, m s⁻¹ for speed).
6. Worked Examples
Modelled example 1
Speed from kinetic energy
Problem
A bullet of mass 15 g has kinetic energy 1200 J. Find its speed.
Study the worked solution
Convert the mass
Method
Express the mass in kilograms.Reason
The kinetic-energy equation requires consistent SI units.Working
15 g = 0.015 kgMake speed the subject
Method
Rearrange Eₖ = (1/2)mv².Reason
The square root is needed because speed is squared in kinetic energy.Working
v = square root of (2Eₖ/m)Calculate the speed
Reason
Substitute the energy and converted mass.Working
v = square root of (2(1200)/0.015) = 400 m s⁻¹
Guided practice 2
Block on a frictionless slope
Problem
A 5.0 kg block slides from rest down a frictionless slope. The distance along the slope is 30 m and the vertical height drop is 5.0 m. Take g = 10 N kg⁻¹. Find its kinetic energy at the bottom.
Choose the relevant height and energy balance
Hints
Hint 1: choose the height
Hint 2: use conservation
View solution step by step
Calculate the initial GPE
Method
Use the vertical height drop.Reason
Gravitational potential energy depends on vertical position, not path length.Working
Eₚ = mgh = (5.0)(10)(5.0) = 250 JTransfer GPE to KE
Reason
The block starts from rest and the frictionless model has no dissipated-energy term.Working
E_(k,bottom) = E_(p,initial) = 250 J
Common misconception 3
Effect of halving speed on kinetic energy
Learner response
A car slows from speed v to v/2. A student says: Its kinetic energy also halves because its speed halves. Locate the first error, find the final kinetic energy as a fraction of the initial value Eᵢ, and find the percentage decrease.
Test the proportional reasoning
View solution step by step
Apply the squared speed dependence
Method
Substitute v/2 into the kinetic-energy equation.Reason
Kinetic energy is proportional to v², not v.Working
E_f = (1/2)m(v/2)² = (1/4)((1/2)mv²) = (1/4)EᵢFind the decrease
Reason
Three quarters of the initial energy is no longer in the car’s kinetic store.Working
Eᵢ-E_f = Eᵢ-(1/4)Eᵢ = (3/4)Eᵢ = 75% decrease
Examiner practice 4
Drop height from landing speed
Examination question
An object of mass 4.0 kg is dropped from rest and reaches the ground with speed 20 m s⁻¹. Take g = 10 N kg⁻¹ and neglect air resistance. Find its landing kinetic energy, its decrease in gravitational potential energy, and the drop height. [4 marks]
Show the energy model and all three results
View solution step by step
Calculate the landing KE
1 markMethod
Use the measured landing speed.Reason
The object has kinetic energy at the ground.Working
Eₖ = (1/2)mv² = (1/2)(4.0)(20²) = 800 JApply conservation
1 markMethod
Equate the GPE decrease to the KE increase.Reason
The object starts from rest and air resistance is neglected.Working
Δ Eₚ = Δ Eₖ = 800 JRelate GPE to height
1 markMethod
Use mgh = 800 J.Reason
The GPE decrease is set by the vertical drop.Working
h = 800/(4.0)(10)State the drop height
1 markReason
The mass and field strength are already in SI units.Working
h = 20 m
Self-mark with the mark scheme
Compare your response with each mark point. Select a point only when your response contains that evidence.
Self-mark the landing KE, conservation step, height substitution and final height separately.
Challenge 5
Block on a rough slope (with energy dissipated)
Changed boundary condition
A 5.0 kg block slides from rest down the same slope through a vertical height of 5.0 m, but the slope is now rough and 30 J is dissipated by friction. Take g = 10 N kg⁻¹. Find its kinetic energy at the bottom.
Add the dissipated-energy term
Hints
Hint 1: find the available energy
Hint 2: include dissipation
View solution step by step
Calculate the initial GPE
Method
Find the energy available before the descent.Reason
The block starts from rest at the same vertical height.Working
Eₚ = (5.0)(10)(5.0) = 250 JInclude the dissipated energy
Method
Write the full energy balance.Reason
Friction transfers 30 J from the block’s mechanical energy stores to internal energy stores.Working
Eₚ = Eₖ + E_dissipatedCalculate the remaining KE
Reason
The final kinetic energy is the initial GPE minus the dissipated energy.Working
Eₖ = 250-30 = 220 J
7. Mind Stretchers
Mind stretcher 1: Speed with energy dissipationExtension
A 1.5 kg block slides down a slope with a vertical height drop of 2.0 m. During the motion, 6.0 J is dissipated due to friction. Take g = 10 N kg⁻¹. Find the speed at the bottom.
Show Answer
Initial GPE: Eₚ = mgh = (1.5)(10)(2.0) = 30 J
Final KE: Eₖ = 30 - 6.0 = 24 J
Eₖ = (1/2)mv²; 24 = (1/2)(1.5)v²; v = square root of (2(24)/1.5) = 5.66 m s⁻¹
Mind stretcher 2: Maximum height with energy dissipatedExtension
A 0.20 kg ball is thrown vertically upwards from the ground with speed 10 m s⁻¹. Before it reaches maximum height, 4.0 J is dissipated due to air resistance. Take g = 10 N kg⁻¹. Find the maximum height reached.
Show Answer
Initial KE: E_(k,initial) = (1/2)mv² = (1/2)(0.20)(10²) = 10 J
At maximum height, Eₖ = 0, so: 10 = mgh + 4.0 ⇒ mgh = 6.0
h = 6.0/(0.20)(10) = 3.0 m
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Course and syllabus information
- Course
- SEC G3 Physics
- Edition
- SEC G3 Physics 2027