Kinetic Energy Calculator

Kinetic energy is the energy an object has because it is moving:

$KE = \tfrac{1}{2}mv^2$

Symbols and SI units:

• $KE$ — kinetic energy, joules (J), where $1\ \text{J} = 1\ \text{kg·m}^2/\text{s}^2$
• $m$ — mass, kilograms (kg)
• $v$ — speed, metres per second (m/s)

Rearranged for the speed, $v = \sqrt{2\,KE/m}$, and for the mass, $m = 2\,KE/v^2$.

Kinetic energy is a scalar: it has no direction and is never negative. It also scales with the square of the speed, so doubling the speed quadruples the energy — the reason stopping distance grows so steeply with speed.

When it applies: translational motion of a point mass or a body moving without spinning. A rotating body carries $\tfrac{1}{2}I\omega^2$ as well.

The assumption people forget: $\tfrac{1}{2}mv^2$ is the non-relativistic form, accurate only while $v \ll c$. Above roughly $0.1c$ the relativistic expression is needed.

The net work done on an object equals the change in its kinetic energy:

$W_{\text{net}} = \Delta KE = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2$

with $W$ in joules, and for a constant force along a straight path $W_{\text{net}} = F_{\text{net}}d$. Combining the two gives the working form used for braking, collisions and accelerating problems:

$F_{\text{net}}\,d = \tfrac{1}{2}m\left(v_f^2 - v_i^2\right)$

This is the fastest route whenever a problem gives you speeds and a distance but no time — you never have to find the acceleration.

Sign: positive net work speeds an object up, negative net work (friction, braking) slows it down.

When it applies: it holds for any force, constant or varying, conservative or not, as long as $W_{\text{net}}$ counts every force acting.

The assumption people forget: $\Delta KE$ responds to the net work only. Work done by one force is cancelled if another does equal and opposite work.

• Squaring only part of the expression — it is $\tfrac{1}{2}m v^2$, so the whole speed is squared before the halving, not after.
• Doubling the energy when you double the speed — the $v^2$ makes it four times larger.
• Using weight instead of mass — $m$ must be in kilograms; a weight in newtons has to be divided by $g$ first.
• Leaving the speed in km/h — divide by $3.6$ to reach m/s before squaring, or the answer is out by a factor of about $13$.
• Giving kinetic energy a sign or a direction — it is a positive scalar even when the velocity is negative.
• Applying $W = \Delta KE$ to one force only — sum the work of every force, friction included, before equating.
• Confusing it with momentum — momentum is $mv$ in kg·m/s and is a vector; the two are not interchangeable.