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.

  • SEC G3 Physics 2027
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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.

Need the concepts first?

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

  1. Mark the initial and final states, including the vertical height change.
  2. List the energy stores at each state and any energy dissipated.
  3. Write the conservation equation in words before using formulas.
  4. 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.

Energy stores during one pendulum swingA pendulum moves from the left turning point A through the lowest point B towards the right. A and an ideal turning point D are at equal height. A real turning point is lower because some energy has been dissipated. Energy bars show all gravitational potential energy at A, all kinetic energy at B, and a mixture of gravitational potential and dissipated energy at the real turning point.Pendulum: choose the lowest point as zero GPEAt a turning point, speed and KE are momentarily zero.AreleaseBlowest pointDidealD′ realsame heightzero-GPE levelGPEKEdissipatedAGPEBKED′GPEdiss.
Scroll diagram horizontally to read all labels.
At A, the bob is momentarily at rest. With negligible resistance, its GPE decrease equals its KE increase at B and it reaches the same height at D. A real pendulum reaches the lower turning point D′ because energy has been dissipated.

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.
Energy accounting for a bouncing ballFour equal-length stacked bars show energy at release, just before impact, just after rebound, and at the rebound's highest point. With negligible air resistance, all initial gravitational potential energy becomes kinetic energy before impact. The collision dissipates part of that energy, leaving equal smaller kinetic and rebound gravitational stores. Each total remains constant.Bouncing ball: total energy is conservedAir resistance negligible; schematic shares—not measured dataGPEKEdissipated to surroundings1 Releasedhighest initial positionGPE2 Before impactfastest downward motionKE3 After reboundcollision has dissipated energyKEdissipated4 Rebound toplower than release heightGPEdissipated
Scroll diagram horizontally to read all labels.
Every bar has the same total length because energy is conserved. With air resistance neglected, the dissipated share increases only during the collision; the smaller rebound GPE corresponds to a smaller rebound height.

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

Core

Problem

A bullet of mass 15 g has kinetic energy 1200 J. Find its speed.

Study the worked solution
  1. Convert the mass

    Method

    Express the mass in kilograms.

    Reason

    The kinetic-energy equation requires consistent SI units.

    Working

    15 g = 0.015 kg
  2. Make 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)
  3. 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

About 5 min

Problem

Choosing height and accounting for energy on a slopeA 5 kilogram block descends a 30 metre slope through a vertical height of 5 metres. Beside it, equal-length energy bars compare a smooth slope, where 250 joules of gravitational potential energy becomes 250 joules of kinetic energy, with a rough slope, where 30 joules is dissipated and 220 joules remains kinetic energy.Slope question: geometry first, then energy accountingWhich distance belongs in mgh?5.0 kgslope distance = 30 mh = 5.0 mΔGPE = mgΔh = 250 JOnly the vertical height change affects GPE.Where does the 250 J go?KEdissipatedSmooth sloperesistance negligible250 J KERough slope30 J dissipated220 J KE30250 J = 220 J + 30 J
Scroll diagram horizontally to read all labels.
Use the 5.0 m vertical height change in mgh, not the 30 m distance along the slope. A smooth slope gives 250 J of KE; dissipating 30 J leaves 220 J of KE.

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

Unit: J
Unit: J

Hints

Hint 1: choose the height
GPE uses vertical height change, so use 5.0 m.
Hint 2: use conservation
Frictionless means no energy-dissipation term in this model.
View solution step by step
  1. 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 J
  2. Transfer 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

Find and correct the mistake

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

Final kinetic energy
Unit: %

View solution step by step
  1. 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)Eᵢ
  2. Find 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

4 marks

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
  1. Calculate the landing KE

    1 mark

    Method

    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 J
  2. Apply conservation

    1 mark

    Method

    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 J
  3. Relate GPE to height

    1 mark

    Method

    Use mgh = 800 J.

    Reason

    The GPE decrease is set by the vertical drop.

    Working

    h = 800/(4.0)(10)
  4. State the drop height

    1 mark

    Reason

    The mass and field strength are already in SI units.

    Working

    h = 20 m

Challenge 5

Block on a rough slope (with energy dissipated)

Minimal support

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

Unit: J

Hints

Hint 1: find the available energy
The initial GPE is still mgh = (5.0)(10)(5.0).
Hint 2: include dissipation
Write Eₚ = Eₖ + E_dissipated.
View solution step by step
  1. 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 J
  2. Include 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_dissipated
  3. Calculate 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