Energy Calculations: KE, GPE and Conservation

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.

2. Key Ideas

• Core equations (O Level):
  • kinetic energy: Eₖ = 1/2mv²
  • 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: gtokg and cmtom.
• 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 = 1/2mv² Rightarrow v = sqrt2gh

The mass cancels because, under these assumptions, both the GPE decrease and KE increase are proportional to mass.

A Level extension (optional)

The derivation of Eₖ = 1/2mv² is not required at O Level. If you want it, see Kinetic Energy From Work Done.

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.

Scroll diagram horizontally to read all labels.

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,

Eₚ,A = Eₖ,B

From point B to point D,

Eₖ,B = Eₚ,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′,

Eₚ,A = Eₚ,D' + Edᵢₛₛᵢₚₐₜₑd, 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.

Scroll diagram horizontally to read all labels.

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 (gtokg and cmtom).
• 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

Example 1: Speed from kinetic energyCore

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

Show Answer

Convert mass: 15 g = 0.015 kg

Eₖ = 1/2mv²; v = sqrt2Eₖ/m = sqrt2(1200)/0.015 = 400 m s⁻¹

Example 2: Effect of halving speed on kinetic energyCore

A car moves at speed v. Its speed decreases to v/2. By what factor does its kinetic energy change? How much does it decrease by (in terms of the initial kinetic energy)?

Show Answer

Let Eᵢ be the initial kinetic energy: Eᵢ = 1/2mv²

Final kinetic energy: Ef = 1/2m(v/2)² = 1/4(1/2mv²) = 1/4Eᵢ

So the kinetic energy becomes one-quarter of the original.
Decrease in kinetic energy: Eᵢ - Ef = Eᵢ - 1/4Eᵢ = 3/4Eᵢ (75\% decrease)

Example 3: Where is GPE maximum and minimum for a pendulum?Core

For the ideal pendulum positions A, B and D in the diagram, at which point or points is the gravitational potential energy (a) maximum and (b) minimum? Take B as the zero-GPE reference.

Show Answer

GPE is Eₚ = mgh, so it is largest at the greatest height.

a) Maximum GPE: at A and D, the equal-height turning points. b) Minimum GPE: at B, the lowest point.

Example 4: Drop height from landing speedCore

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:

1. its kinetic energy as it lands,
2. its decrease in gravitational potential energy, and
3. the drop height.

Show Answer

1. Kinetic energy on landing: Eₖ = 1/2mv² = 1/2(4.0)(20²) = 800 J

2. With no initial KE and negligible resistance: decrease in GPE = increase in KE = 800 J

3. Using mgh = 800: h = 800/(4.0)(10) = 20 m

Example 5: Gain in GPE when liftingCore

An object of mass 5.0 kg is lifted vertically through 10 m at constant speed. Take g = 10 N kg⁻¹. Find the gain in gravitational potential energy.

Show Answer

Δ Eₚ = mgh = (5.0)(10)(10) = 500 J

Example 6: Hammering a nailCore

A raised hammer is used to drive a nail into a plank. Describe the energy changes.

Show Answer

• Raised hammer: gravitational potential energy store.
• Falling hammer: its GPE store decreases while its KE store increases.
• Impact: energy is transferred mechanically to the nail and plank. Their internal energy stores increase, and sound waves carry some energy away.

Example 7: Throwing and catching a ballCore

A student throws a ball into the air and catches it on the way down. Describe the changes in its energy stores and the final transfer.

Show Answer

• As the ball rises, its KE store decreases while its GPE store increases.
• At the highest point, KE is momentarily zero and GPE is maximum.
• As it falls, its GPE store decreases while its KE store increases.
• When it is caught, energy is transferred mechanically from the ball. Internal energy stores increase, and sound waves carry some energy away.

Example 8: Block on a frictionless slopeCore

Scroll diagram horizontally to read all labels.

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 (from the diagram). Take g = 10 N kg⁻¹. Find its kinetic energy at the bottom.

Show Answer

Initial GPE: Eₚ = mgh = (5.0)(10)(5.0) = 250 J

Frictionless ⇒ no energy dissipated, so final KE = initial GPE: Eₖ = 250 J

Example 9: Block on a rough slope (with energy dissipated)Core

The same 5.0 kg block slides down the slope, but 30 J is dissipated due to friction. Find its kinetic energy at the bottom.

Show Answer

From Example 8, initial GPE = 250 J.

Conservation of energy: Eₚ = Eₖ + Edᵢₛₛᵢₚₐₜₑd

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 = 1/2mv²; 24 = 1/2(1.5)v²; v = sqrt2(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: Eₖ,ᵢₙᵢₜᵢₐₗ = 1/2mv² = 1/2(0.20)(10²) = 10 J

At maximum height, Eₖ = 0, so: 10 = mgh + 4.0 Rightarrow mgh = 6.0

h = 6.0/(0.20)(10) = 3.0 m

8. Practice, Quiz and Next Step

Close your notes and use Energy Calculations: KE, GPE and Conservation in the supplied context below. This requires a constructed explanation or working, not recognition of an option.

Fresh context: A 2.0 kg load is raised vertically by 5.0 m and later moves at 6.0 m/s; use g = 10 N/kg.

1. Retrieve: define kinetic and gravitational potential energy in your own words, including units, sign or conditions where relevant.
2. Represent: Draw separate energy-store diagrams for the raised and moving states.
3. Apply: Calculate the gravitational potential and kinetic energy, then explain why equality cannot be assumed without a stated transfer process.

Check the response before looking back

• The chosen system and transfer pathway are stated before applying conservation.
• Energy and power are not used interchangeably, and all quantities use compatible units.
• The numerical result is checked against the physical situation and any efficiency limit.

If one check fails, name that exact gap, revisit the matching explanation or worked example, and redo the task with different values or a different situation. Then use theO-Level topic checks orpractice browser for an independent re-test.