Kinetic Energy Calculator Eₖ = ½mv²: Work, Speed-Squared Scaling and Energy Units

About this calculator

Kinetic energy is the energy associated with motion. For a mass $m$ moving at speed $v$ , the classical non-relativistic formula is

E_k = \frac12 mv^2.

Enter mass and speed to calculate energy, or enter energy with one of the other variables to solve for the missing quantity. Because speed is squared, doubling speed multiplies kinetic energy by four; doubling mass only doubles it.

This model is appropriate for ordinary macroscopic speeds. It is not a relativistic calculator and should not be used near the speed of light.

Formula and variables

E_k = \frac12 mv^2,\qquad v = \sqrt{\frac{2E_k}{m}},\qquad m = \frac{2E_k}{v^2}

Symbol	Meaning	SI unit	Notes
$m$	mass	kg	use mass, not weight
$v$	speed	m/s	magnitude of velocity
$E_k$	kinetic energy	J	$1\,J=1\,kg\,m^2/s^2$

Useful conversions include $1\,kWh=3.6\times10^6\,J$ and $1\,eV\approx1.602176634\times10^{-19}\,J$ .

How to use this calculator

Open the kinetic-energy calculator in the Math/Science tools.
Enter mass in kilograms and speed in meters per second to obtain energy in joules.
To solve for speed, clear speed and provide mass plus kinetic energy.
To solve for mass, clear mass and provide speed plus kinetic energy.
Keep units consistent before entering values; convert km/h to m/s by dividing by 3.6.

Worked examples

Example 1: 1,500 kg car at 20 m/s

At 40 m/s, the same car has $1.2\,MJ$ , four times as much energy.

Example 2: 0.45 kg ball at 30 m/s

Example 3: solve speed from 10 kJ and 80 kg

Change speed from 20 to 40 while keeping mass fixed. The result should increase by a factor of four because kinetic energy scales with $v^2$ .

math

E_k=\frac12(1500)(20)^2=300000\,J=300\,kJ.

math

E_k=\frac12(0.45)(30)^2=202.5\,J.

math

v=\sqrt{\frac{2(10000)}{80}}=\sqrt{250}\approx15.8113883\,m/s.

Principle and derivation

The work-energy theorem states that net work changes kinetic energy. For one-dimensional motion with constant mass,

W=\int F\,dx=\int ma\,dx.

Using $a=v\,dv/dx$ gives

W=\int mv\,dv=\frac12m(v_2^2-v_1^2)=\Delta E_k.

That derivation explains why the speed is squared and why braking energy rises so quickly with speed.

Assumptions

Classical mechanics is valid; speeds are far below the speed of light.
Mass is constant during the calculation.
Speed is a scalar magnitude; direction does not affect kinetic energy.
SI units are used by the calculator fields.

Limitations

Supported:: solve any one of $E_k$ , $m$ , and $v$ from the other two.
Approximate only:: real braking and collisions distribute energy into heat, deformation, sound, and other channels; air resistance also adds losses.
Not covered:: rotational kinetic energy, relativistic kinetic energy, total energy of multi-object systems, collision velocity allocation, or potential-energy bookkeeping.
No unit conversion:: convert km/h, grams, kWh, kcal, and other units before entering values.
Numerics:: negative kinetic energy has no real classical solution for speed; extreme values may use scientific notation.

Common pitfalls

Entering km/h as if it were m/s.
Using weight in newtons instead of mass in kilograms.
Forgetting the square on speed.
Assuming twice the speed means twice the braking energy; it means four times the energy.

Typical use cases

Transport safety

Compare kinetic energy and approximate braking demand at different speeds, including the square-law effect of speed.

Sports and equipment

Estimate the translational energy of balls, projectiles, or moving exercise equipment from mass and speed.

Machinery and energy

Reason about impact energy, moving masses, or the translational part of stored mechanical energy.

Teaching

Introduce the work-energy theorem, square-law scaling, joules, and the contrast between scalar energy and vector momentum.

FAQ

Is kinetic energy ever negative?

No. In the classical formula, mass is positive and $v^2$ is non-negative, so kinetic energy is non-negative.

Does direction matter?

No. Reversing velocity changes momentum sign but not kinetic energy because $(-v)^2=v^2$ .

Why can stopping distance grow roughly with speed squared?

If the available braking force is roughly constant, the work needed to remove kinetic energy is force × distance, while kinetic energy grows as $v^2$ .

Privacy

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Sources & review

Reviewed by: CalcX 编辑组
Updated: 2026-09-09

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