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All lessons Mechanics23 min

Conservation of Energy

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01
Hook
02
Explore
03
Formalize
04
Practice
05
Challenge
Interactive simulation
01

Hook

A roller-coaster car is hauled to the top of the first hill, then released — no engine, no fuel, nobody pushing. It plunges faster and faster, whips through a loop, and climbs the next hill almost as high as it started. Where does all that speed come from?

02

Explore

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03

Formalize

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04

Practice

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05

Challenge

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Spoilers

Conservation of Energy — summary and key formula

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The question

A roller-coaster car is hauled to the top of the first hill, then released — no engine, no fuel, nobody pushing. It plunges faster and faster, whips through a loop, and climbs the next hill almost as high as it started. Where does all that speed come from?

The energy was quietly stored in the car's height the whole time it was being hauled up. Watching height turn into speed and back reveals one rule: energy is never created or destroyed, only transformed — which lets you predict the speed of anything falling from any height, no stopwatch needed.

The key idea

The total mechanical energy of an object is its kinetic energy (motion) plus its gravitational potential energy (height). With no friction or air resistance, this total stays perfectly constant — energy only shifts back and forth between the two forms.

Because the total is fixed, every gain in speed is paid for by an equal loss in height. Equating two points — 12mv12+mgh1=12mv22+mgh2\tfrac{1}{2}mv_1^2 + mgh_1 = \tfrac{1}{2}mv_2^2 + mgh_221​mv12​+mgh1​=21​mv22​+mgh2​ — gives the speed at any height without knowing how long, steep, or curved the path was. And setting all the PE at the top equal to all the KE at the bottom, mgh = ½mv², the mass divides out, leaving v = √(2gh): a 1 kg and a 100 kg ball dropped from the same height arrive at the same speed — exactly Galileo's result. Real systems lose a little energy to friction and air as heat, falling just short of this ideal. **Connect it:** conservation of energy is the work–energy theorem with gravity doing the work: mghmghmgh and 12mv2\tfrac{1}{2}mv^221​mv2 are the same joules in different costumes, and friction's 'loss' is just a transfer into heat you stopped tracking. Follow the joules and every mechanics problem becomes accounting.

The formula

12mv12+mgh1=12mv22+mgh2\frac{1}{2}mv_1^2 + mgh_1 = \frac{1}{2}mv_2^2 + mgh_221​mv12​+mgh1​=21​mv22​+mgh2​
  • ·KE = kinetic energy (J); PE = gravitational potential energy (J); m = mass (kg); v = speed (m/s); g ≈ 9.8 m/s²; h = height above a chosen reference (m). Put h = 0 wherever convenient — only height differences matter.

Common mistake

Forgetting that friction removes mechanical energy — without friction KE + PE is constant, but friction turns mechanical energy into heat so the total drops.

What to remember

  • ·Without friction, ½mv² + mgh stays constant — energy swaps between KE and PE.
  • ·Speed at the bottom of a frictionless drop is independent of mass.
  • ·Friction converts mechanical energy to heat, lowering the total.