# Kinetic energy

Kinetic energy is energy that a body has as a result of its speed.

It is formally defined as work needed to accelerate a body from rest to a velocity v. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work would also be required to return the body to a state of rest from that velocity.

# Simple explanation

Energy can exist in many forms, for example chemical energy, heat, electromagnetic radiation, potential energy (both gravitational and elastic), nuclear energy and kinetic energy.

These forms of energy can often be converted to other forms. Kinetic energy can be best understood by examples that demonstrate how it is transformed from other forms of energy and to the other forms. For example a cyclist will use chemical energy that was provided by food to accelerate a bicycle to a chosen velocity. This velocity can be maintained without further work, except to overcome air-resistance and friction. The energy has been converted into the energy of motion, known as kinetic energy but the process is not completely efficient and heat is also produced within the cyclist.

The kinetic energy in the moving rat and the cyclist can be converted to other forms. For example, the cyclist could encounter a hill just high enough to coast up, so that the bicycle comes to a complete halt at the top. The kinetic energy has now largely been converted to gravitational potential energy that can be released by freewheeling down the other side of the hill. (There are some frictional losses so that the bicycle will never quite regain all the original speed.) Alternatively the cyclist could connect a dynamo to one of the wheels and also generate some electrical energy on the descent. The bicycle would be travelling more slowly at the bottom of the hill because some of the energy has been diverted into making electrical power. Another possibility would be for the cyclist to apply the brakes, in which case the kinetic energy would be dissipated as heat energy.

## Simple calculation

The kinetic energy of a body = $\displaystyle \begin{matrix} \frac{1}{2} \end{matrix} mv^2$

where m is the mass and v is the velocity of the body.

Note that the kinetic energy increases with the square of the velocity. This means for example that if you are travelling twice as fast, you need to lose four times as much energy to stop. The braking distance of a car at 100 km/h is therefore four times as far as the braking distance at 50 km/h.

## More simple examples

The space shuttle uses chemical energy to take off and gains considerable kinetic energy because it must reach orbital velocity. This kinetic energy gained during launch will remain constant while the shuttle is in orbit because there is almost no friction. However it becomes apparent at re-entry when the kinetic energy is converted to heat.

Kinetic energy can be passed from one object to another. In the game of billiards, the player gives kinetic energy to the cue ball by striking it with the cue stick. If the cue ball collides with another ball, it will slow down dramatically and the ball it collided with will accelerate to a velocity as the kinetic energy is passed on to it. Collisions in billiards are elastic collisions, where kinetic energy is preserved.

Flywheels are being developed as a method of energy storage see article flywheel energy storage. This illustrates that kinetic energy can also be rotational. Note the formula in the articles on flywheels for calculating rotational kinetic energy is different, though analogous.

# Rigorous definitions

$\displaystyle E_k = \int \mathbf{F} \cdot \mathrm{d}\mathbf{s} = \int \mathbf{v} \cdot \mathrm{d}\mathbf{p} = (1/2)mv^2$

the words in the above equation state that the kinetic energy (Ek) is equal to the integral of the dot product of the velocity (v) of a body and the infinitesimal of the body's momentum (p).

## In Newtonian mechanics

For non-relativistic mechanics, the total kinetic energy of a body can be considered as the sum of the body's translational kinetic energy and its rotational energy, or angular kinetic energy:

$\displaystyle E_k = E_t + E_r \,$

where:

Ek is the total kinetic energy
Et is the translational kinetic energy
Er is the rotational energy or angular kinetic energy

For the translational kinetic energy of a body with constant mass m, whose centre of mass is moving in a straight line with linear velocity v, as seen above is equal to

$\displaystyle E_t = \begin{matrix} \frac{1}{2} \end{matrix} mv^2$

where:

m is mass of the body
v is linear velocity of the centre of mass body

Thus, for a speed of 10 m/s the kinetic energy is 50 J/kg, for a speed of 10 km/s it is 50 MJ/kg, etc.

If a body is rotating, its rotational kinetic energy or angular kinetic energy is simply sum of kinetic energies of its moving parts, and thus is equal to:

$\displaystyle E_r = \begin{matrix} \frac{1}{2} \end{matrix} I \omega^2$

where:

I is the body's moment of inertia
ω is the body's angular velocity.

The kinetic energy of a system depends on the inertial frame of reference. It is lowest with respect to the center of mass, i.e., in a frame of reference in which the center of mass is stationary. In another frame of reference the additional kinetic energy is that corresponding to the total mass and the speed of the center of mass.

## In Relativistic mechanics

In Einstein's relativistic mechanics, (used especially for near-light velocities) mass no longer stays constant and accurate calculation of work to accelerate body results in the following expression for kinetic energy:

$\displaystyle E_k = m c^2 (\gamma - 1) = \gamma m c^2 - m c^2 \;\!$
$\displaystyle \gamma = \frac{1}{\sqrt{1 - (v/c)^2}}$
$\displaystyle E_k = \gamma m c^2 - m c^2 = \left( \frac{1}{\sqrt{1- v^2/c^2 }} - 1 \right) m c^2$

where:

Ek is the kinetic energy of the body
v is the velocity of the body
m is its rest mass
c is the speed of light in a vacuum
γmc2 is the total energy of the body
mc2 is the rest mass energy (90 petajoule/kg)

It is an edifying exercise to show that the ratio of this relativistic kinetic energy to the Newtonian kinetic energy given by (1/2)mv2 approaches 1 as v approaches 0, i.e.,

$\displaystyle \lim_{v\to 0}{\left( \frac{1}{\sqrt{1- v^2/c^2\ }} - 1 \right) m c^2 \over mv^2/2}=1$

This can be done by the techniques of first-year calculus.

Relativity theory states that the kinetic energy of an object grows towards infinity as its velocity approaches the speed of light, and thus that it is impossible to accelerate an object to this boundary.

Where gravity is weak, and objects move at much slower velocities than light (e.g. in everyday phenomena on Earth), Newton's formula is an excellent approximation of relativistic kinetic energy.

The next term in the approximation is 0.375 mv4/c², e.g. for a speed of 10 km/s this is 0.04 J/kg, for a speed of 100 km/s it is 40 J/kg, etc.

The exact Taylor series is

$\displaystyle E_k = {1\over 2}mv^2 + {3\over 8}m\left({v^4\over c^2}\right) + {5\over 16}m\left({v^6\over c^4}\right) + \dots \,$