Kinetic Energy Calculator

Enter the mass in kilograms and the speed in meters per second. The calculator finds classical translational kinetic energy from Ek = ½mv² and also reports momentum, the speed in km/h or mph, and the energy at double speed. For 1,000 kg at 20 m/s the result is 200,000 J. The model does not include a relativistic correction or rotational energy ½Iω².

Inputs

Result

Enter a mass and a speed. The energy, the momentum, and the Ek(v) chart will appear here.

Work of a force along a distance is on the W = F·s calculator, stopping distance is on the decelerated-motion calculator, and rotational energy is on the E = ½Iω² calculator.

How it works

How Ek is computed for 1,000 kg at 20 m/s
v²400 m²/s²
× m1,000 kg
× ½200,000
Ek200,000 J
m v Eₖ = ½·m·v²
Energy grows with the square of speed: twice the speed is four times the energy.

Translational kinetic energy is the energy a body has because its center of mass is moving. The formula Ek = ½mv² is classical: square the speed, multiply by the mass, and divide by two. The unit is the joule, 1 J = 1 N·m = 1 kg·m²/s². For 1,000 kg at 20 m/s: v² = 400 m²/s², and ½·1,000·400 = 200,000 J. At 40 m/s the same mass has 800,000 J.

The calculator uses mass m in kilograms and speed v in meters per second. Enter 72 km/h as 20, not as 72. About 50 km/h is 13.9 m/s. A comma and a period are the same decimal separator. The sign of the speed does not change the result, because the formula uses v². A negative mass is rejected. Zero mass or zero speed gives 0 J.

What the result panel shows. Besides the energy in a chosen unit (J, kJ, MJ), the calculator reports the same value in joules and watt-hours, the square v², the momentum p = m|v|, the speed in m/s and km/h (or in ft/s and mph after the header switch), the mass you entered, and the energy at double speed. For 1,000 kg at 20 m/s the momentum is 20,000 kg·m/s, and 200,000 J is 55.56 Wh. The chart draws the parabola Ek(v) from one quarter of the entered speed to twice that speed, when mass and speed are greater than zero.

Limits of the model. The formula is translational motion at constant mass and at speeds far below the speed of light. The calculator does not add the rotational energy ½Iω², does not compute potential energy, does not apply (γ − 1)mc², and does not report a stopping distance. Work from force and distance is on the W = F·s page, and distance at a known deceleration is on the decelerated-motion page. Momentum as its own quantity is on the p = mv page.

The header switch converts mass to pounds, speed to ft/s, and energy to ft·lb, and it converts numbers already entered in the fields. It does not turn kilometers per hour into meters per second by itself: do that conversion before you enter the speed. The numbers in the text stay in SI units as the reference.

The examples on the page use the same formula: 1,000 kg at 20 m/s gives 200,000 J, an 8 g bullet at 350 m/s gives 490 J, an 80 kg sprinter at 10 m/s gives 4,000 J, and classical ½mv² for an Earth mass at 29,780 m/s is on the order of 2.648×10³³ J and is not the total orbital energy. Fill this example enters those same numbers into the fields and refreshes the panel and the chart.

Formula

Ek = ½ · m · v²

[Ek] = kg · (m/s)² = kg·m²/s² = J; p = mv; Ek = p²/(2m)

Model: classical translational kinetic energy (Leibniz’s vis viva, 1686; the factor ½ from G.-G. Coriolis, 1829). The rearrangements m = 2Ek/v² and v = √(2Ek/m) follow from the same formula, but this calculator solves only for Ek. Order of calculation: v², then ½mv² in SI units, then p = m|v|.

Theory, derivation, and scope

Definition and the joule in the SI

The translational kinetic energy of a body of mass m and speed v is Ek = ½mv². It is the energy stored in the motion of the center of mass, not a separate kind of energy. The SI unit is the joule: 1 J = 1 N·m = 1 kg·m²/s². The name honors James Prescott Joule. The unit is defined in the SI Brochure (BIPM, 9th edition).

For 1,000 kg at 20 m/s the steps are short. Step 1: v² = 20² = 400 m²/s². Step 2: m·v² = 1,000·400 = 400,000 kg·m²/s². Step 3: half of that product is 200,000 J. The same mass at 40 m/s has v² = 1,600 and an energy of 800,000 J. A mass of 2,000 kg at 20 m/s has 400,000 J, because mass enters the formula linearly.

History: Huygens, Leibniz, and the factor ½

In De motu corporum ex percussione, written about 1656 and published posthumously in 1703, Christiaan Huygens showed that an elastic collision conserves the sum of mass times speed squared. In “Brevis demonstratio erroris memorabilis Cartesii”, published in Acta Eruditorum in March 1686, Gottfried Wilhelm Leibniz set mv², which he called living force (vis viva), against the Cartesian measure mv. The vis viva dispute ran through the eighteenth century.

Émilie du Châtelet defended mv² in Institutions de physique (1740). She cited the 1722 clay experiments of ’sGravesande, in which the depth of the mark grew with the square of the speed. In Du calcul de l’effet des machines (1829), Gaspard-Gustave Coriolis introduced the factor ½ so that the change in ½mv² would equal the work of a force. William Thomson and Peter Guthrie Tait spread the name kinetic energy for that quantity in Treatise on Natural Philosophy (1867). In 1807 Thomas Young used the word energy for a quantity proportional to mv², still without the factor ½. This page does not invent DOIs, because these works predate that system.

The work–energy theorem

For a particle of constant mass, Newton’s second law integrated along the path gives the work–energy theorem. Net work equals the change in kinetic energy: W = ½mv₂² − ½mv₁². Starting from rest, the work required to reach speed v is exactly ½mv². For 1,000 kg at 20 m/s that is 200,000 J.

The calculator reports that store of energy. It has no force field and no distance field, so it does not turn joules into meters. If a constant force of 2,000 N accelerated that mass from rest to 20 m/s, the distance would be 100 m, because 200,000 J / 2,000 N = 100 m. That illustrates the theorem. It is not an output of this form. Distance at a known deceleration belongs on the decelerated-motion page.

Momentum, the square of speed, and numeric examples

Momentum p = mv is linear in speed and has the unit kg·m/s. The energy can be written Ek = p²/(2m). For 1,000 kg at 20 m/s the momentum is 20,000 kg·m/s, and p²/(2m) = 400,000,000 / 2,000 = 200,000 J, in agreement with the panel. Doubling the speed doubles the momentum and quadruples the energy. Doubling the mass at the same speed doubles both.

A 1,400 kg car at 13.9 m/s, about 50 km/h, has 135,247 J. At 33.3 m/s, about 120 km/h, it has 776,223 J. An 8 g bullet at 350 m/s has 490 J. That is the kinetic energy of the center of mass, without the chemical energy of the charge and without spin about the bullet’s axis. An 80 kg sprinter at 10 m/s has 4,000 J. A 50,000 kg train at 20 m/s has 10 MJ.

The order of magnitude for Earth in orbit about the Sun uses a mass of about 5.972×10²⁴ kg and a mean orbital speed of about 29.78 km/s, or 29,780 m/s, from the NASA planetary fact sheet. Classical ½mv² is then about 2.648×10³³ J. In Newtonian mechanics the total energy of a circular orbit is negative and equal in magnitude to this kinetic energy, because the potential energy is twice as large and negative. The calculator does not compute the gravitational constant, the semimajor axis, or that difference. It also omits the reduced mass, a correction on the order of the Earth-to-Sun mass ratio, about 3×10⁻⁶.

The relativistic limit

Relativistic kinetic energy is (γ − 1)mc², where γ = 1/√(1 − v²/c²) and c = 299,792,458 m/s. The low-speed expansion begins with ½mv². At 20 m/s the relative difference is on the order of 10⁻¹⁵ and does not change any digit of the result. One tenth of the speed of light already leaves the classical value about 0.75% low. Half the speed of light leaves it about 19% low, and at 0.8c the classical result is about 48% of the relativistic kinetic energy.

The calculator always returns ½mv². It does not multiply by γ and it has no relativistic mode. When the entered speed is at least 0.1c, the note beside the result recalls this limit, but the number on the panel stays classical. For Earth’s orbit the relative correction is on the order of 10⁻⁸ and is not computed either.

What this formula does not compute

The formula does not include rotation. A rolling wheel also has ½Iω², and the share of that term depends on the shape of the body. The sum belongs on the rotational kinetic energy page, not here. The formula does not include friction, air resistance, deformation, or the heat of an inelastic collision. It also does not include the chemical energy of a cartridge: the 490 J of the bullet is the energy the bullet already carries, not the energy of the propellant.

The formula is not orbital energy. The Earth example shows the order of magnitude of ½mv² and nothing more. It does not replace celestial mechanics or relativistic dynamics.

Sources

The definition, the unit, and the work–energy theorem follow mechanics textbooks and the historical sources: D. Halliday, R. Resnick, and J. Walker, Fundamentals of Physics, chapters on work and kinetic energy; H. D. Young and R. A. Freedman, University Physics, the chapter on work and kinetic energy; R. P. Feynman, R. B. Leighton, and M. Sands, The Feynman Lectures on Physics, Volume I, chapters 4 and 13; C. Huygens, De motu corporum ex percussione, 1703 edition; G. W. Leibniz, Acta Eruditorum, March 1686; G.-G. Coriolis, Du calcul de l’effet des machines, 1829; W. Thomson and P. G. Tait, Treatise on Natural Philosophy, 1867; BIPM, SI Brochure, 9th edition; NASA, Planetary Fact Sheet (Earth mass and orbital speed). The page is educational.

How to use it

  1. Enter the mass in kilograms, for example 1,000. After you switch the header to US units, the field accepts pounds.
  2. Enter the speed in meters per second, for example 20. 72 km/h is 20, not 72. About 50 km/h is 13.9.
  3. Choose Calculate, or wait for the live result. For 1,000 kg at 20 m/s the calculator shows 200 kJ.
  4. Check the panel: joules, watt-hours, v², momentum, and the energy at double speed. The chart is the parabola Ek(v).
  5. Clearing the form removes the numbers you entered and hides the panel. The examples enter ready mass–speed pairs, including a bullet, a sprinter, and an order of magnitude for Earth’s orbit.

Symbol glossary

Ek is translational kinetic energy, m is mass, and v is speed. For 1,000 kg at 20 m/s the result is 200,000 J.

Ek
Translational kinetic energy [J], Ek = ½mv². The panel also shows kJ and Wh.
m
Mass [kg]. Zero gives 0 J. A negative value is rejected.
v
Speed [m/s]. 72 km/h = 20 m/s. The sign does not change Ek, because the formula uses v².
J
Joule, 1 J = 1 kg·m²/s². 200,000 J = 55.56 Wh, because 1 Wh = 3,600 J.
p
Momentum [kg·m/s], p = m|v|. With it, Ek = p²/(2m). It is a different quantity from energy.
½
Coriolis’s factor from 1829, so that the change in Ek equals work.
γ
The Lorentz factor. (γ − 1)mc² is relativistic kinetic energy, which this calculator does not compute.

Worked examples

Example 1

A 1,000 kg car at 20 m/s, which is exactly 72 km/h and about 45 mph. This is the reference case on the page: ½·1,000·400 = 200,000 J.

Ek = 200,000 J

Example 2

The same 1,000 kg mass at 40 m/s. The speed is twice as large, and the energy is four times as large, because v is squared.

Ek = 800,000 J

Example 3

A family car of 1,400 kg at 13.9 m/s, about 50 km/h or 31 mph. That is the center-of-mass energy the brakes have to remove for the car to stop.

Ek = 135,247 J

Example 4

The same 1,400 kg car at 33.3 m/s, about 120 km/h or 75 mph. The speed is a little more than twice the city figure, and the energy is nearly six times larger.

Ek = 776,223 J

Example 5

A bullet of 8 g, or 0.008 kg, at 350 m/s. The 490 J result is the kinetic energy of the center of mass, not the chemical energy of the cartridge and not the spin of the bullet.

Ek = 490 J

Example 6

A sprinter of 80 kg at 10 m/s, which is 36 km/h or about 22 mph. That is a running speed, not a walking step, and the energy is exactly 4,000 J.

Ek = 4,000 J

Example 7

An order of magnitude for Earth: a mass of about 5.972×10²⁴ kg and an orbital speed about the Sun of 29,780 m/s. Classical ½mv² gives about 2.648×10³³ J. This is not the total orbital energy, which is negative in Newtonian mechanics, and it is not a relativistic correction.

Ek ≈ 2.648×10³³ J

Example 8

A cyclist and bicycle together, 85 kg, at 6.9 m/s, about 25 km/h or 15 mph. The center-of-mass energy is 2,023 J. Wheel rotation is a separate ½Iω² term.

Ek = 2,023 J

Example 9

A simplified passenger train of 50,000 kg at 20 m/s, or 72 km/h. The energy is 10,000,000 J, or 10 MJ, which is why a train stopping distance is measured in hundreds of meters.

Ek = 10 MJ

Example 10

A match ball of 0.43 kg at 25 m/s, or 90 km/h. The center-of-mass energy is 134.4 J. This formula leaves out the spin of the ball.

Ek = 134.4 J

Example 11

A motorcycle of 200 kg at 25 m/s, or 90 km/h. Less mass than a car, and the energy is still 62,500 J.

Ek = 62,500 J

Example 12

Twice the mass at the same speed: 2,000 kg at 20 m/s. Energy grows linearly with mass and is 400,000 J, not four times larger.

Ek = 400,000 J

Common questions

What does this calculator compute, and from which formula?

The calculator finds the classical translational kinetic energy from Ek = ½mv². It squares the speed, multiplies by the mass, and divides by two. For 1,000 kg at 20 m/s the result is 200,000 J.

Which units should I enter for mass and speed?

Enter mass in kilograms and speed in meters per second. The result is in joules. 72 km/h is 20 m/s, not the number 72 in the speed field. After you switch the header to US units, mass is in pounds, speed is in ft/s, and energy is in ft·lb. A comma and a period are the same decimal separator.

Why does doubling the speed give four times the energy?

Because speed is squared in the formula. Twice the speed is four times v², so four times the energy. 1,000 kg at 40 m/s is 800,000 J, not 400,000 J. The result panel shows that value on its own. Ten times the speed gives one hundred times the energy.

How does the result relate to the work–energy theorem?

For a body of constant mass, the net work equals the change in kinetic energy: W = ΔEk = ½mv₂² − ½mv₁². Starting from rest, the work required to reach speed v is exactly ½mv². The calculator reports that store of energy. It does not compute a force, a distance, or a stopping distance.

Does the calculator apply a relativistic correction?

No. The result is always the classical ½mv². Relativistic kinetic energy is (γ − 1)mc², where γ = 1/√(1 − v²/c²), and this calculator does not compute it. At 20 m/s the difference is on the order of 10⁻¹⁵ relative to the result. At speeds comparable to the speed of light the classical formula understates the energy. The page states this as a limit of the model, not as a second calculation mode.

How is translational energy different from ½Iω²?

½mv² describes motion of the center of mass. ½Iω² describes rotation about an axis and lives on the separate rotational kinetic energy page. A rolling wheel has both terms, and this calculator does not add them. Momentum p = mv is a different quantity, linear in v, in kg·m/s. They are linked by Ek = p²/(2m).

What do zero and a minus sign mean?

A negative mass is rejected. Zero mass or zero speed gives 0 J, because there is no translational kinetic energy. The sign of the speed does not change the result, because the formula uses v². Both fields need a number before the panel appears.

What do the panel and the chart show?

The panel reports the energy, the same value in joules and watt-hours, the square of the speed, the momentum p = m|v|, the speed in m/s and km/h (or in ft/s and mph after the header is switched to US units), the mass you entered, and the energy at double speed. The chart draws the parabola Ek(v) at constant mass, from one quarter of the speed to twice that speed, when both values are greater than zero. The Fill this example button enters the example numbers and refreshes the panel and the chart. The fields start empty.

Is the Earth-orbit example the energy of the orbit?

No. The example computes classical ½mv² for an Earth mass of about 5.972×10²⁴ kg and an orbital speed of 29,780 m/s. The result is on the order of 2.648×10³³ J. That is the kinetic energy of the center of mass, not the total orbital energy, which is negative in Newtonian mechanics. The calculator does not apply the virial theorem, a reduced mass, or the factor γ.

Does the result include friction, deformation, or air resistance?

No. The result is the energy of motion at one instant. Friction and drag reduce the speed over time, and an inelastic collision turns part of the energy into heat and deformation. The calculator does not subtract those losses. It does not replace a collision analysis or a stopping distance.

How do I calculate kinetic energy with Ek = ½mv²?

Square the speed, multiply by the mass, and divide by two. For 1,000 kg at 20 m/s: ½·1,000·400 = 200,000 J.

What is Ek for 1,000 kg at 20 m/s?

200,000 J, or 200 kJ. At 40 m/s the same mass has 800,000 J, because speed is squared.

How do I convert km/h or mph into m/s before the formula?

Divide km/h by 3.6, or multiply mph by 0.44704. 72 km/h is 20 m/s; about 45 mph is also 20 m/s. Do not leave a km/h or mph number in the speed field without converting.

Does the calculator solve for mass or speed from a given energy?

Not with this field layout. You enter mass and speed, and the result is Ek plus momentum. The rearranged forms m = 2Ek/v² and v = √(2Ek/m) are not separate modes on this page.

How is Ek = ½mv² different from potential energy mgh?

½mv² is the energy of center-of-mass motion. mgh is gravitational potential energy in a uniform field. This calculator only finds the translational kinetic term.

The result is half the product Ek = ½mv²: for 1,000 kg at 20 m/s it is 200,000 J, and the momentum p = mv is 20,000 kg·m/s. Rotational energy is on the E = ½Iω² calculator.

Sources

The definition, the unit, the history, and the work–energy theorem come from physics and metrology sources. The page does not invent publication identifiers.

Page updated November 3, 2026.