Kinetic Energy Calculator

Calculate kinetic energy using KE = ½mv². Enter mass and velocity to find energy in joules, kJ, or kWh. Includes speed doubling examples and crash physics.

Kinetic energy is the energy an object possesses due to its motion. The faster an object moves, or the heavier it is, the more kinetic energy it carries — and the more energy is required to stop it.

Kinetic Energy Formula

KE = ½ × m × v²

Where:

  • KE = Kinetic energy in joules (J)
  • m = Mass in kilograms (kg)
  • v = Velocity in metres per second (m/s)

Example: A 1,000 kg car travelling at 14 m/s (50 km/h / 31 mph):

KE = 0.5 × 1,000 × 14² = 0.5 × 1,000 × 196 = 98,000 J = 98 kJ

Why Velocity Matters More Than Mass

The quadratic relationship between speed and kinetic energy is one of the most important concepts in crash physics — and everyday driving.

Because kinetic energy uses v², doubling speed quadruples kinetic energy:

SpeedKE (1,000 kg car)Relative to 30 mph
30 mph (13.4 m/s)89.8 kJ
60 mph (26.8 m/s)359 kJ
90 mph (40.2 m/s)808 kJ
120 mph (53.6 m/s)1,436 kJ16×

A car at 60 mph doesn't just need twice the braking distance of a car at 30 mph — it needs four times the distance. This is the physics behind speed limit enforcement near schools and pedestrian zones.

Kinetic Energy in Real Scenarios

ObjectMassSpeedKinetic Energy
Tennis ball (serve)0.058 kg70 m/s (252 km/h)142 J
70 kg cyclist90 kg total8 m/s (29 km/h)2,880 J
1,000 kg car1,000 kg14 m/s (50 km/h)98,000 J (98 kJ)
40-tonne lorry40,000 kg25 m/s (90 km/h)12,500,000 J (12.5 MJ)
Boeing 747 landing300,000 kg70 m/s (252 km/h)735,000,000 J (735 MJ)
Meteorite (small)1,000 kg20,000 m/s200,000,000,000 J (200 GJ)

Energy Unit Conversions

UnitEquivalentWhen used
1 joule (J)1 kg·m²/s²Physics calculations
1 kilojoule (kJ)1,000 JCars, machinery
1 megajoule (MJ)1,000,000 JLarge vehicles, explosions
1 kilowatt-hour (kWh)3,600,000 JElectricity billing
1 calorie (cal)4.184 JChemistry (not nutrition)
1 kilocalorie (kcal)4,184 JFood/nutrition labelling
1 BTU1,055 JUS heating/cooling

Example: A 1,500 kg car at 100 km/h has KE = 0.5 × 1,500 × 27.78² = 578,750 J ≈ 579 kJ ≈ 0.161 kWh.

That's the same energy as running a 160W light bulb for one hour — consumed entirely in the braking process.

The Work-Energy Theorem

Kinetic energy and the work done by forces are directly related:

Work done (J) = Change in kinetic energy (J)
W = KE_final − KE_initial

Braking example: A 1,200 kg car brakes from 60 mph (26.8 m/s) to 0:

KE_initial = 0.5 × 1,200 × 26.8² = 430,752 J
KE_final = 0
Work done by brakes = −430,752 J

The brakes must absorb 430 kJ — this energy becomes heat in the brake discs and pads.

Kinetic Energy vs Potential Energy

Kinetic energy (motion) and potential energy (position/height) convert into each other in conservative systems:

Potential energy: PE = m × g × h
(where h = height in metres, g = 9.81 m/s²)

A 70 kg person at the top of a 10-metre diving board:

PE = 70 × 9.81 × 10 = 6,867 J

At the water surface (h = 0), all potential energy has converted to kinetic energy:

KE = 6,867 J → v = √(2 × KE ÷ m) = √(2 × 6,867 ÷ 70) = 14 m/s (50.4 km/h)

Frequently Asked Questions

What happens to kinetic energy when an object stops?

Kinetic energy is converted into other forms. In braking: heat in brake pads and discs. In a collision: deformation energy (crumpling metal), sound, and heat. In friction: heat. Energy is never destroyed — it changes form, per conservation of energy.

Does kinetic energy depend on direction?

No. Kinetic energy is a scalar quantity — it has magnitude but no direction. A car moving north at 60 mph and a car moving east at 60 mph have the same kinetic energy, even though their velocities (vector quantities) are different.

Why is there a ½ in the kinetic energy formula?

The ½ comes from the calculus derivation. Force = ma, and kinetic energy is the work done accelerating from rest: KE = ∫F·dx. Substituting F = ma and using the kinematic relationship v² = 2as gives KE = ½mv². The ½ is exact, not an approximation.

What is rotational kinetic energy?

Objects can also have rotational kinetic energy from spinning: KE_rot = ½ × I × ω², where I is the moment of inertia (kg·m²) and ω is angular velocity (rad/s). A spinning flywheel or a rolling tyre has both translational and rotational kinetic energy.

How does aerodynamic drag relate to kinetic energy?

Aerodynamic drag increases with the square of speed (F_drag ∝ v²). Combined with kinetic energy also increasing as v², fuel consumption from drag increases as v³ — which is why driving at 80 mph uses dramatically more fuel than 60 mph. Reducing speed from 80 to 70 mph cuts drag-related fuel consumption by about 30%.

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