3.4.1.8MechanicsCore

Conservation of Energy

The $principle of conservation of energy$ is one of the deepest ideas in physics: energy is never created or destroyed, only transferred between stores. In mechanics this lets you swap between gravitational potential and kinetic energy without ever touching the forces — and it neatly accounts for the energy “lost” to friction as heat.

40 min Video by Physics Online 3.4.1 Force, energy and momentum

Conservation of Energy — A-Level PhysicsWatch the full walkthrough before the notes below.

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What you'll be able to do

State the principle of conservation of energy
Track energy transfers between kinetic, gravitational and thermal stores
Solve problems by equating energy before and after a transfer
Account for work done against resistive forces as dissipated energy
Recognise why total energy is conserved even when mechanical energy is not

The principle of conservation of energy

The principle of conservation of energy states that the total energy of a closed system stays constant: energy cannot be created or destroyed, only $transferred$ from one store to another or carried away by heating, radiation, sound and so on.

This is a book-keeping law. If you add up every store before an event and every store after it, the totals must match. Any energy that seems to “disappear” has simply been transferred somewhere you haven’t counted yet — most often to a thermal store.

Tip — Start energy problems by naming the stores involved before and after. “Where did the energy come from, and where did it go?” is often faster than a force-and-acceleration approach.

Interchange of kinetic and potential energy

When resistive forces are negligible, gravitational potential energy and kinetic energy convert freely into each other while their $total stays constant$ . A falling object loses $m g Δ h$ of potential energy and gains exactly that much kinetic energy.

Setting the two equal, $m g Δ h = \frac{1}{2} m v^{2}$ , the mass cancels — so, ignoring air resistance, every object dropped through the same height reaches the same speed regardless of how heavy it is.

m g Δ h = \frac{1}{2} m v^{2} \Rightarrow v = 2 g Δ h

Speed gained by falling a height Δh with no resistive losses.

1All the lost gravitational PE becomes kinetic energy:

m g Δ h = \frac{1}{2} m v^{2}

2The mass cancels, leaving

v = 2 g Δ h = 2 \times 9.81 \times 1.8

v = 35.3 = 5.9

m/s.

Answer

v \approx 5.9

m/s

Work done against resistive forces

Real systems lose mechanical energy to friction, drag and air resistance. That energy is not destroyed — it is transferred to a $thermal$ store, warming the surfaces and surroundings. The work done against the resistive force equals the mechanical energy dissipated.

So the full energy balance for something sliding or falling with resistance is: initial energy = final energy + energy dissipated. A roller coaster reaching a lower second hill than the first is conservation of energy in action, not a violation of it.

E_{start} = E_{end} + W_{against resistance}

Mechanical energy lost equals the work done against resistive forces (transferred to heat).

1GPE available

= m g Δ h = 60 \times 9.81 \times 40 = 23, 544

2KE gained

= \frac{1}{2} m v^{2} = \frac{1}{2} (60) (25)^{2} = 18, 750

3Energy dissipated

= 23, 544 - 18, 750 \approx 4.8 \times 1 0^{3}

Answer

\approx 4.8 \times 1 0^{3}

J transferred to heat

Tip — If a moving object ends up slower than a resistance-free calculation predicts, the missing kinetic energy has gone to a thermal store via work done against friction.

Equation recap

E_{total, before} = E_{total, after}

Conservation of energy for a closed system.

m g Δ h = \frac{1}{2} m v^{2}

PE ↔ KE interchange with no losses.

v = 2 g Δ h

Speed after falling Δh (no resistance).

E_{start} = E_{end} + W_{resistance}

Energy balance with dissipation.

Common mistakes to avoid

Thinking energy is destroyed when an object slows down and stops.

Energy is only transferred — the kinetic energy becomes thermal energy through work done against friction.

Keeping the mass in a PE-to-KE calculation and expecting heavier objects to fall faster.

In mgΔh = ½mv² the mass cancels, so (ignoring air resistance) the landing speed is independent of mass.

Ignoring resistive losses when a problem clearly involves friction or drag.

Include a dissipation term: initial energy = final energy + work done against resistance.

Using the slope length instead of the vertical height for gravitational PE.

ΔE_p = mgΔh uses the vertical drop Δh, not the distance travelled along the slope.

Key takeaways

Energy cannot be created or destroyed, only transferred between stores.
With no resistance, gravitational PE and kinetic energy interchange with a constant total: mgΔh = ½mv².
The landing speed from a height is v = √(2gΔh), independent of mass.
Work done against friction and drag transfers mechanical energy to a thermal store — total energy is still conserved.

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