Calculate kinetic energy from mass and velocity, or work backwards to find speed or mass. Get results in joules, kJ, kWh, calories, and foot-pounds. Three calculation modes, real-world examples, and the work-energy theorem — everything you need in one place.
Enter mass and velocity — choose your preferred units. Results appear in joules, kJ, kWh, calories, and foot-pounds.
Know the kinetic energy and mass? Find the object’s speed using v = √(2 × KE / m).
Know the kinetic energy and speed? Find the object’s mass using m = 2 × KE / v².
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Pick up any physics textbook and you’ll find the same deceptively simple equation near the beginning: KE = ½mv². Half the mass times the square of the velocity. It takes about three seconds to write and a lifetime to fully appreciate. Why? Because the velocity is squared — which means speed changes energy in a way that feels almost unfair. Double your speed and you don’t double your kinetic energy. You quadruple it. That single fact explains everything from why car crashes at 60 mph are far deadlier than crashes at 30 mph, to why a baseball thrown at 90 mph carries so much more force than one lobbed at 45 mph.
The full form of the formula is KE = ½ × m × v², where mass m is in kilograms and velocity v is in meters per second. The result is in joules (J), named after James Prescott Joule, who established the mechanical equivalent of heat in the 1840s. The factor of one-half comes directly from integrating Newton’s second law (F = ma) over distance — it’s not arbitrary; it’s a mathematical necessity.
What this formula is really telling you is how much work was needed to accelerate an object from rest to its current speed. Flip that around: it’s also exactly how much work an object at that speed can do against a braking force before it stops. That’s why engineers care deeply about kinetic energy when designing brakes, crash structures, and anything else that has to absorb motion.
KE = 0.5 x m x v^2 (joules, J)
v = sqrt(2 x KE / m) (m/s, given KE in J)
m = 2 x KE / v^2 (kg, given KE in J)
W = delta-KE = KE_f - KE_i (work-energy theorem)
1 J = 0.7376 ft-lb = 0.2390 cal = 2.778e-4 Wh = 6.242e18 eV
Double speed: KE x 4 | Triple speed: KE x 9
Theory is only useful when you can apply it. Here are three worked examples showing how to use the formula with different objects and units — each one highlighting a different aspect of the calculation.
Example 1 — a car at 60 km/h: A 1,500 kg car travels at 60 km/h. First, convert to m/s: 60 ÷ 3.6 = 16.67 m/s. Then: KE = 0.5 × 1,500 × 16.67² = 0.5 × 1,500 × 278.0 = 208,500 J = 208.5 kJ. That’s about the same energy as lifting the car from the ground to the top of a five-storey building.
Example 2 — a baseball pitch: A professional pitcher throws a 0.145 kg baseball at 40 m/s (about 89 mph): KE = 0.5 × 0.145 × 40² = 0.5 × 0.145 × 1,600 = 116 J = 85.6 ft-lb. It doesn’t sound like much in joules, but concentrated into a 70 mm sphere travelling at that speed, the impact energy is significant enough to cause serious injury.
Example 3 — finding velocity from KE: A 4 kg object has 800 J of kinetic energy. What is its speed? v = √(2 × 800 / 4) = √400 = 20 m/s (72 km/h). This reverse calculation is exactly what engineers do when estimating impact speeds from crash damage evidence.
One of the best ways to build intuition for kinetic energy is to compare familiar objects. The numbers below might surprise you — especially how dramatically speed changes energy compared to mass.
| Object | Mass | Speed | KE (Joules) | KE (ft-lb) | Context |
|---|---|---|---|---|---|
| Person walking | 70 kg | 1.4 m/s (5 km/h) | 69 J | 51 ft-lb | Typical adult walking pace |
| Baseball pitch | 0.145 kg | 40 m/s (89 mph) | 116 J | 86 ft-lb | Professional MLB fastball |
| Car @ 60 km/h | 1,500 kg | 16.7 m/s | 208,500 J | 153,800 ft-lb | Urban speed limit typical |
| Car @ 120 km/h | 1,500 kg | 33.3 m/s | 834,000 J | 615,200 ft-lb | 4× more than at 60 km/h |
| 9mm bullet | 0.0075 kg | 370 m/s | 514 J | 379 ft-lb | Standard 9×19 mm Parabellum |
| .308 rifle round | 0.0097 kg | 838 m/s | 3,408 J | 2,513 ft-lb | NATO 7.62×51 mm at muzzle |
| Freight train | 6,000,000 kg | 27.8 m/s (100 km/h) | 2,317,200,000 J | 1.71 billion ft-lb | Why stopping trains is so hard |
| Skydiver terminal | 80 kg | 55 m/s (200 km/h) | 121,000 J | 89,240 ft-lb | Without parachute |
Here’s the insight that catches most people off guard: mass and velocity do not contribute equally to kinetic energy. Double the mass and KE doubles. But double the speed and KE quadruples. Triple the speed? Nine times the energy. This is why road safety campaigns focus so heavily on speed reduction rather than weight reduction — lowering a car’s speed by 20% reduces kinetic energy by 36%, which is a far bigger safety gain than any feasible weight reduction.
The same principle applies in ballistics. A lighter bullet fired at higher velocity can carry more kinetic energy than a heavier, slower round. In engineering design, materials must be rated to absorb the peak kinetic energies they might encounter — and the squared velocity term means even modest overspeed scenarios require dramatically stronger structures.
The work-energy theorem ties together force, displacement, and kinetic energy in a single elegant statement: the net work done on an object equals the change in its kinetic energy. Written as W = ΔKE = KEfinal − KEinitial. This is one of the most useful relations in all of mechanics because it sidesteps the need to solve equations of motion — you only need to know the starting and ending speeds.
Practical applications are everywhere. A car’s braking system converts kinetic energy into heat through friction. An electric motor adds kinetic energy to a flywheel. A parachute does negative work on a falling skydiver, reducing kinetic energy and slowing the descent. A spring launcher does positive work on a ball, increasing its kinetic energy. Every time energy is transferred to or from motion, the work-energy theorem is at work.
Physics uses joules (SI), but engineers, ballisticians, nutritionists, and electrical engineers each have their preferred energy units. Here’s what you need to know for the most common conversions. The calculator above outputs all of these simultaneously so you never have to convert manually: