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Conservation of Mechanical Energy (½mv² + mgh)

Also known as conservation of energy · mechanical energy conservation · energy conservation ramp · speed from height · roller coaster energy · KE plus PE constant · half mv squared plus mgh

12v12+gh1=12v22+gh2\tfrac{1}{2} v_1^{2} + g h_1 = \tfrac{1}{2} v_2^{2} + g h_2

Worked example: Dropped from rest at 10 m → 14.0047 m/s at the bottom — press Try an example to run it live, then adjust anything.

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Constant used — built into this formula, no need to enter
gn=9.80665 m/s2g_{n} = 9.80665\ \text{m/s}^{2}Standard acceleration of gravity · exact
See also
Kinetic Energy — The ½mv² term on its own, when the energy itself rather than the speed is what you need.Gravitational Potential Energy (U = mgh) — The mgh term on its own, and where the height's reference level comes from.Work–Energy Theorem — What to reach for the moment friction enters: energy stops being conserved and the lost work has to be accounted for.

Learning zone

On a frictionless path, kinetic energy and gravitational potential energy trade back and forth but their sum never changes: ½mv² + mgh is the same at the top of the hill as at the bottom. Write that out at two points, divide every term by the mass, and the mass vanishes — which is the single most useful thing this equation says. A 2 kg cart and a 200 kg cart released from the same 10 m drop both arrive at √(2 × 9.80665 × 10) ≈ 14.0 m/s. Galileo's inclined planes were this result being measured three centuries before anyone wrote it in these symbols.

The other thing it says is that the shape of the path is irrelevant. A ramp, a loop, a corkscrew and a vertical drop of the same height all deliver the same speed, because gravity does the same work on the way down whatever route the track takes. That is why a roller coaster's first hill has to be the tallest, and why a pendulum bob returns to precisely the height it was released from and no higher. The word doing all the work in the setup is frictionless. Real track, real air and real bearings take a cut, so treat the answer as the ceiling: measure less and the difference is the energy that went into heat. Once that loss matters, the balance stops being conservation and becomes the work–energy theorem, where the lost work has to be named and subtracted.

Conservation of Mechanical Energy (½mv² + mgh)
12v12+gh1=12v22+gh2\tfrac{1}{2} v_1^{2} + g h_1 = \tfrac{1}{2} v_2^{2} + g h_2
Where
  • v1v_1= Speed at point 1 (m/s)
  • h1h_1= Height at point 1 (m)
  • v2v_2= Speed at point 2 (m/s)
  • h2h_2= Height at point 2 (m)
Missing one of these? Work it out first, then come back