Conservation of energy on a roller coaster

TL;DR:

• Without friction or drag, mgh at the top turns into ½mv² at the bottom, so v = √(2gh).

• The mass appears on both sides and cancels. A heavy train and a light one reach the same speed.

• Real rides lose energy to friction and air, so the true speed is lower than the ideal one.

A coaster train creeps up the first hill, tips over the top, and then plunges. You do not need to know the shape of the track to find its speed at the bottom. Energy conservation lets you skip the curve entirely and compare just two moments: the top and the bottom.

The idea in plain words

Energy is not created or destroyed, only moved between forms. For a coaster car there are two that matter here. Gravitational potential energy depends on height: PE = mgh. Kinetic energy depends on speed: KE = ½mv².

If only gravity does work on the car, the sum KE + PE stays the same at every point. Falling lowers PE and raises KE by the same amount. Climbing does the reverse.

Why the path does not matter

Gravity is a conservative force: the work it does depends only on the change in height, not the route. A straight ramp, a spiral and a twisty track with the same drop give the same energy change. That is why a roller-coaster designer can set the speed at any point from its height alone, before drawing any curves.

The normal force from the track pushes perpendicular to the motion, so it does no work. That is why it also drops out of the energy equation.

Worked example: the speed at the bottom of a 20 m drop

The train crests a hill 20 m above the bottom, moving so slowly we treat it as starting from rest. Ignore friction and air. Use g = 9.8 m/s².

1. Write the conservation statement: KE_top + PE_top = KE_bottom + PE_bottom.

2. Choose the bottom as height zero. At the top: 0 + mgh. At the bottom: ½mv² + 0.

3. Set equal: mgh = ½mv². The mass cancels.

4. Solve: v² = 2gh = 2 × 9.8 × 20 = 392, so v = √392 ≈ 19.8 m/s.

5. In everyday units: 19.8 × 3.6 ≈ 71 km/h.

Now ask how high the train could climb on the far side. At the bottom its kinetic energy per kilogram is ½v² = ½ × 392 = 196 J/kg. Climbing to height h costs gh per kilogram, so h = 196 / 9.8 = 20 m. The train can rise back to exactly its starting height and no higher.

That is the rule behind coaster design. Each later hill must be lower than the first, because some energy is always lost. A hill 15 m high, for instance, would be crossed at v = √(2 × 9.8 × (20 − 15)) = √98 ≈ 9.9 m/s in the ideal case.

Why mass cancels, and what it means

Mass sits in both mgh and ½mv². A heavier train has more energy to convert, but it also has more inertia to speed up, and the two effects match exactly. This is the same reason a heavy and a light ball dropped together hit the ground together. If your answer depends on the mass, you have made an algebra slip.

A bar chart makes the trade easy to see. At the top, all 196 J/kg is potential. Halfway down, at 10 m, half is potential (9.8 × 10 = 98 J/kg) and half is kinetic, so v = √(2 × 98) = 14 m/s. At the bottom, all of it is kinetic. The total bar height never changes, only the split between the two colours.

How to check your answer

• Units. √(m/s² × m) = √(m²/s²) = m/s.

• Substitute back. ½ × 19.8² = ½ × 392 = 196 per kg, and g × h = 9.8 × 20 = 196 per kg. Equal.

• Free-fall comparison. A stone dropped from 20 m reaches the same 19.8 m/s. The track only steers the motion; it does not change the speed.

• Upper bound. The real speed should be a bit lower, never higher, than 19.8 m/s.

What real losses do

Friction in the wheels and drag from the air turn some mechanical energy into heat. Take a hypothetical measurement of 18 m/s at the bottom. Kinetic energy per kg is ½ × 18² = 162 J/kg, compared with the ideal 196 J/kg. The missing 196 − 162 = 34 J/kg went into heat and sound.

Notice friction acts over the whole path length, so a longer, twistier track loses more than a short straight one with the same drop. This is the one place where the shape of the track does matter.

Note: Energy conservation tells you speed, not the force on riders. A tight loop at high speed can still give large forces on the body. That belongs to circular motion, covered in circular motion and centripetal force.

Common mistakes

• Measuring height along the track instead of straight vertically.

• Choosing a different zero level for the top and the bottom.

• Forgetting to square v or taking the square root too early.

• Leaving out the starting speed when the train does not start from rest.

• Treating friction as zero in a problem that gives a friction value.

Frequently asked questions

Does the steepness of the drop change the speed at the bottom?

Not without friction. Only the height difference matters. A steeper drop gets there sooner, but the speed at the bottom is the same.

Why does a real coaster need a motor at the start?

The first lift supplies the energy for the whole ride. Friction and drag take some of it away continuously, so every later hill is lower.