Kinetic energy (SI unit: the joule) is energy that a body possesses as a result of its motion. It is formally defined as the work needed to accelerate a body from rest to its current velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. Negative work of the same magnitude would be required to return the body to a state of rest from that velocity.

In simple cases, this work (and thus the kinetic energy) is equal to:

$E\_k\; =\; \backslash begin\{matrix\}\; \backslash frac\{1\}\{2\}\; \backslash end\{matrix\}\; mv^2$

where m is the object's mass and v is the object's speed.

Origin of term

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The etymology of 'kinetic energy' is the Greek word for motion kinesis and the Greek word for active work energeia. Therefore the term 'kinetic energy' means through motion do active work. The terms kinetic energy and work and their present scientific meanings date back to the mid 19th century. Early understandings of these ideas can be attributed to Gaspard-Gustave Coriolis who in 1829 published the paper titled Du Calcul de l'effet des machines outlining the mathematics of kinetic energy.

Simple explanation

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Energy can exist in many forms, for example chemical energy, heat, electromagnetic radiation, potential energy (gravitational, electric, elastic, etc.), nuclear energy, mass, 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 speed. This speed 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 bicycle 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.

See also energy conversion.

Simple calculation

In classical mechanics, the kinetic energy of a "point object" (a body so small that its size can be ignored) is given by the equation $E\_k\; =\; \backslash begin\{matrix\}\; \backslash frac\{1\}\{2\}\; \backslash end\{matrix\}\; mv^2$ where m is the mass and v is the speed of the body.

For example - one would calculate the kinetic energy of an 80 kg mass travelling at 40 mph (17.8816 metres per second) as $\backslash begin\{matrix\}\; \backslash frac\{1\}\{2\}\; \backslash end\{matrix\}\; *\; 80\; *\; 17.8816^2\; =\; 12790$ Joules.

Note that the kinetic energy increases with the square of the speed. This means for example that if you are traveling twice as fast, you need to lose four times as much energy to stop.

More simple examples

Spacecraft use chemical energy to take off and gain considerable kinetic energy to reach orbital velocity. This kinetic energy gained during launch will remain constant while 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 speed 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.

Definition

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$E\_k\; =\; \backslash int\; \backslash mathbf\{F\}\; \backslash cdot\; \backslash mathrm\{d\}\backslash mathbf\{s\}\; =\; \backslash int\; \backslash mathbf\{v\}\; \backslash cdot\; \backslash mathrm\{d\}\backslash mathbf\{p\}$

This equation states that the kinetic energy (E_k) is equal to the integral of the dot product of the velocity (v) of a body and the infinitesimal change of the body's momentum (p). It is assumed that the body starts at rest (motionless).

In Newtonian mechanics

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For non-relativistic mechanics, the formula above gives:

$E\_k\; =\; \backslash frac\{1\}\{2\}mv^2$

It sometimes is convenient to split the total kinetic energy of body into the sum of the body's center-of-mass translational kinetic energy and the energy of rotation around the center of mass rotational energy:

$E\_k\; =\; E\_t\; +\; E\_r\; \backslash ,$

where:

E_k is the total kinetic energy

E_t is the translational kinetic energy

E_r 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 speed v, as seen above is equal to

$E\_t\; =\; \backslash begin\{matrix\}\; \backslash frac\{1\}\{2\}\; \backslash end\{matrix\}\; mv^2$

where:

m is mass of the body

v is speed of the centre of mass body

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:

$E\_r\; =\; \backslash begin\{matrix\}\; \backslash frac\{1\}\{2\}\; \backslash end\{matrix\}\; I\; \backslash omega^2$

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.

Thus kinetic energy is a relative measure and no object can be said to have a unique kinetic energy. A rocket engine could be seen to transfer its energy to the rocket ship or to the exhaust stream depending upon the chosen frame of reference. But the total energy of the system, ie kinetic energy, fuel chemical energy, heat energy etc, will be conserved regardless of the choice of measurement frame.

The kinetic energy of an object is related to its momentum by the equation:

$E\_k\; =\; \backslash frac\{p^2\}\{2m\}$

In relativistic mechanics

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Einstein's relativistic mechanics must be used for calculating the kinetic energy of bodies whose speeds are a significant fraction of the velocity of light:

Recall Einstein's famous formula:

$E\; =\; mc^2$

For an object in motion,

$m\; =\; \backslash frac\{m\_0\}\{\backslash sqrt\{1\; -\; (v/c)^2\}\}$

where m₀ is the rest mass, v is the object's speed, and c is the speed of light in vacuum.

So

$E\; =\; mc^2\; =\; c^2(\backslash frac\{m\_0\}\{\backslash sqrt\{1\; -\; (v/c)^2\}\})$

The equation shows that the energy of an object approaches infinity as the velocity v approaches the speed of light c, thus it is impossible to accelerate an object across this boundary.

By substituting $x\; =\; (v/c)^2$ we can rewrite this as:

$E\; =\; c^2(m\_0\; (1\; -\; x)^\{-\backslash frac\{1\}\{2\}\})$

The first two Taylor series coefficients of the correction factor $f(x)\; =\; (1\; -\; x)^\{-\backslash frac\{1\}\{2\}\}$ are:

$f(0)\; =\; (1\; -\; 0)^\{-\backslash frac\{1\}\{2\}\}\; =\; 1$

$f\text{'}(0)\; =\; \backslash frac\{1\}\{2\}(1\; -\; 0)^\{-\backslash frac\{3\}\{2\}\}\; =\; \backslash frac\{1\}\{2\}$

So we approximate $f(x)\; \backslash approx\; f(0)\; +\; f\text{'}(0)x$:

$E\; =\; c^2(m\_0\; (1\; -\; x)^\{-1/2\})\; \backslash approx\; c^2\; (m\_0\; (1\; +\; \backslash frac\{1\}\{2\}x))\; =\; c^2\; (m\_0\; (1\; +\; \backslash frac\{1\}\{2\}(v^2/c^2)))\; =\; m\_0\; c^2\; +\; \backslash frac\{1\}\{2\}\; m\_0\; v^2$

Indicating that the total energy can be partitioned into the rest mass's energy plus the traditional newtonian energy (at low speeds).

When objects move at speeds much slower than light (e.g. in everyday phenomena on Earth), the first two terms of the series predominate. The next term in the approximation is small for low speeds, e.g. for a speed of 10 km/s the correction to the Newtonian kinetic energy is 0.04 J/kg and for a speed of 100 km/s it is 40 J/kg, etc.

For higher speeds, the formula for the relativistic kinetic energy is derived by simply subtracting out the rest mass energy.

$E\_k\; =\; mc^2\; -\; m\_0\; c^2\; =\; m\_0\; c^2(\backslash frac\{1\}\{\backslash sqrt\{1\; -\; (v/c)^2\}\}\; -\; 1)$

The relation between kinetic energy and momentum is more complicated in this case, and is given by the equation:

$E\_k\; =\; \backslash sqrt\{p^2c^2+m\_0^2c^4\}-m\_0c^2$

This can also be expanded as a Taylor series, the first term of which is the simple expression from Newtonian mechanics.

what this suggests is that the formulae for energy and momentum are not special and axiomatic, but rather concepts which emerge from the equation of mass with energy and the principles of relativity.

In quantum mechanics

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In quantum wave-mechanics, the expectation value of the electron kinetic energy, , for a system of electrons described by the wavefunction $\backslash vert\backslash psi>$ is a sum of 1-electron operator expectation values (written here in atomic units)

where $\backslash nabla^2\_i$ is the Laplacian operator acting upon the coordinates of the i'th electron and the summation runs over all electrons.

The density functional formalism of quantum mechanics requires knowledge of the electron density only, i.e., it formally does not require knowledge of the wavefunction. Given an electron density $\backslash rho(\backslash mathbf\{r\})$, the exact N-electron kinetic energy functional is unknown; however, for the specific case of a 1-electron system, the kinetic energy can be written as

$T*=\; \backslash int\; \backslash frac\{1\}\{8\}\backslash frac\{\backslash nabla\; \backslash rho(\backslash mathbf\{r\})\; \backslash cdot\; \backslash nabla\; \backslash rho(\backslash mathbf\{r\})\; \}\{\; \backslash rho(\backslash mathbf\{r\})\; \}\; d^3r$

where $T*$ is known as the Weizsacker kinetic energy functional.

See also

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• Recoil
• Joule
• Kinetic projectile
• Potential energy

References

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Energy | Classical mechanics | Introductory physics

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