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Conservation of Energy

The total mechanical energy of an isolated system remains constant. Energy transforms between kinetic and potential forms.

Last updated: 12 February 2026

β˜…β˜…β˜…β˜†β˜† Level 3PhysicsEnergyConservationPotential EnergyKinetic Energy3 min read

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Law of Conservation of Energy

Energy can neither be created nor destroyed; it can only be transformed from one form to another.

For mechanical energy (no friction):

Conservation of Mechanical Energy

Ek+Ep=constantβ€…β€ŠβŸΉβ€…β€Š12mv2+mgh=constE_k + E_p = \text{constant} \implies \frac{1}{2}mv^2 + mgh = \text{const}

Ball dropped from height 20 m. Find speed at 8 m height.

At top: Ek=0E_k = 0, Ep=mg(20)E_p = mg(20)

At h = 8 m: 12mv2+mg(8)=mg(20)\frac{1}{2}mv^2 + mg(8) = mg(20)

v=2gΓ—12=2Γ—9.8Γ—12β‰ˆ15.3Β m/sv = \sqrt{2g \times 12} = \sqrt{2 \times 9.8 \times 12} \approx 15.3 \text{ m/s}

Note

When friction is present, some mechanical energy is converted to thermal energy. The total energy is still conserved, but mechanical energy decreases.

Real world: Ski jump

A 75 kg ski jumper starts from rest at the top of an 80 m ramp (ignoring friction and air resistance).

At the bottom: 12mv2=mghβ€…β€ŠβŸΉβ€…β€Šv=2Γ—10Γ—80=1600=40Β m/s\frac{1}{2}mv^2 = mgh \implies v = \sqrt{2 \times 10 \times 80} = \sqrt{1600} = 40 \text{ m/s} (144 km/h)

In practice, friction and air drag mean actual speeds are lower β€” but the conservation principle tells us the theoretical maximum.

Real world: Pendulum clock

A pendulum bob is released from a height of 0.2 m above its lowest point (g = 10 m/sΒ²).

At the lowest point all GPE has become KE: v=2gh=2Γ—10Γ—0.2=4=2Β m/sv = \sqrt{2gh} = \sqrt{2 \times 10 \times 0.2} = \sqrt{4} = 2 \text{ m/s}

The bob then rises back to exactly 0.2 m on the other side (in an ideal, frictionless system). Real pendulums slowly lose height because air resistance converts mechanical energy to heat.

Real world: Roller coaster loop

A roller coaster car starts at height 45 m and enters a circular loop of radius 10 m (bottom of loop at h = 0).

Speed at the bottom of the loop: v=2Γ—10Γ—45=900=30Β m/sv = \sqrt{2 \times 10 \times 45} = \sqrt{900} = 30 \text{ m/s}

At the top of the loop (h = 20 m): v=2Γ—10Γ—(45βˆ’20)=500β‰ˆ22.4Β m/sv = \sqrt{2 \times 10 \times (45 - 20)} = \sqrt{500} \approx 22.4 \text{ m/s}

Engineers use energy conservation to check the car is still fast enough at the top to stay on the track.

Tip

Where do you see this in real life? Roller coasters, playground swings, bungee jumping, and ski runs all rely on energy being converted between kinetic and potential forms β€” conservation of energy lets engineers design safe speeds at every point on the ride.

Practice Problems

A pendulum bob swings from height 0.4 m. Find its speed at the lowest point.

Show answer
v = √(2 Γ— 9.8 Γ— 0.4) = √7.84 β‰ˆ 2.8 m/s