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In simple cases, this work (and thus the kinetic enrgy) is equal to:
:E_K = rac{mv^2}{2}
where ''m'' is the object's Mass and ''v'' is the object's Speed .


ORIGIN OF TERM


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

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 = \begin{matrix} rac{1}{2} \end{matrix} mv^2 where m is the mass and v is the speed of the body.

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

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 speed as the kinetic energy is passed on to it. Collisions in billiards are elastic collisions, where kinetic energy is preserved.

Flywheel s 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

: E_k = \int \mathbf{F} \cdot \mathrm{d}\mathbf{s} = \int \mathbf{v} \cdot \mathrm{d}\mathbf{p} = rac{1}{2}mv^2

This equation states that the kinetic energy (''Ek'') 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

For non-relativistic mechanics, 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 \,

where:
Ek

Et

Er


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 = \begin{matrix} rac{1}{2} \end{matrix} mv^2

where:
m

v


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 = \begin{matrix} rac{1}{2} \end{matrix} I \omega^2

where:
I

: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.


In relativistic mechanics

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:

v

c





This can be shown using the Taylor Series expansion:

:{(\gamma - 1)mc^2\over {1\over2}mv^2} = \sum_{n=1}^{\infty}\left( rac{(2n-1)!}{n!(n-1)!2^{2n-1}} rac{v^{2n-2}}{c^{2n-2}} ight) = 1 + {3\over 4}\left({v^2\over c^2} ight) + {5\over 8}\left({v^4\over c^4} ight) + \cdots.

When objects move at speeds much slower than light (e.g. in everyday phenomena on Earth), the first term of the series predominates. 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.


SEE ALSO



REFERENCES