Lasted edited by Andrew Munsey, updated on June 14, 2016 at 9:21 pm.

- 34 errors has been found on this page. Administrator will correct this soon.
- This page has been imported from the old peswiki website. This message will be removed once updated.

: See also Directory:Kinetic Energy Devices

Kinetic energy (SI unit: the `There was an error working with the wiki: Code[1]`

needed to accelerate a body from rest to its current velocity. Having gained this energy during its `There was an error working with the wiki: Code[9]`

, the body maintains this kinetic energy unless its speed changes. `There was an error working with the wiki: Code[10]`

work of the same magnitude would be required to return the body to a state of rest from that velocity. The etymology of 'kinetic energy' is the Greek word for motion `There was an error working with the wiki: Code[11]`

and the Greek word for active work `There was an error working with the wiki: Code[12]`

. 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 `There was an error working with the wiki: Code[13]`

who in 1829 published the paper titled Du Calcul de l'effet des machines outlining the mathematics of kinetic energy.

: E_k = \int {F} \cdot {d}{s} = \int {v} \cdot {d}{p}

This equation states that the kinetic energy (Ek) is equal to the `There was an error working with the wiki: Code[14]`

of the `There was an error working with the wiki: Code[15]`

of the Velocity (v) of a body and the `There was an error working with the wiki: Code[16]`

change of the body's `There was an error working with the wiki: Code[17]`

(p). It is assumed that the body starts at rest (motionless).

The following holds only if we assume that we are dealing with Newtonian mechanics and that we have a point object. Simply writing "in the non-relativistic case" doesn't get the job done, and elaborating the necessary conditions is too burdensome here. Under certain assumptions, this work (and thus the kinetic energy) is equal to:

:E_k = \frac{1}{2} mv^2

where m is the object's Mass and v is the object's `There was an error working with the wiki: Code[18]`

.

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. Various 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.

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.

`There was an error working with the wiki: Code[19]`

is any process of converting energy from one form to another. Energy found in fossil fuels, solar radiation, or nuclear fuels needs to be converted into other energy forms such as electrical, propulsive, or cooling to be useful. Machines are used to convert energy from one form to another. The efficiency of a machine characterizes how well it can convert the energy from one form to another. Energy is converted so that it may be used by other machines or to provide an energy service to society. An internal combustion engine converts the chemical energy in the gasoline to the propulsive energy that moves a car. A solar cell converts the solar radiation into the electrical energy that can then be used to light a bulb or power a computer.

In `There was an error working with the wiki: Code[20]`

, the kinetic energy of a "point object" (a body so small that its size can be ignored) is given by the equation E_k = \frac{1}{2} 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 \frac{1}{2} \cdot 80 \cdot 17.8816^2 = 6395 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.

For non-relativistic mechanics, the formula above gives:

:E_k = \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 `There was an error working with the wiki: Code[21]`

:

: 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 `There was an error working with the wiki: Code[22]`

is moving in a straight line with speed v, as seen above is equal to

: E_t = \frac{1}{2} mv^2

where:

:m is mass of the body

:v is speed of the `There was an error working with the wiki: Code[23]`

body

If a body is rotating, its `There was an error working with the wiki: Code[2]`

or angular kinetic energy is simply sum of kinetic energies of its moving parts, and thus is equal to:

: E_r = \frac{1}{2} I \omega^2

where:

:I is the body's `There was an error working with the wiki: Code[24]`

:? is the body's `There was an error working with the wiki: Code[25]`

.

The kinetic energy of a system depends on the `There was an error working with the wiki: Code[26]`

. It is lowest with respect to the `There was an error working with the wiki: Code[27]`

, 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 `There was an error working with the wiki: Code[28]`

by the equation:

:E_k = \frac{p^2}{2m}

`There was an error working with the wiki: Code[3]`

's `There was an error working with the wiki: Code[4]`

must be used for calculating the kinetic energy of bodies whose speeds are a significant fraction of the velocity of light. As Einstein's formula states:

: E = mc^2.

For an object in motion:

: m = \frac{m_0}{\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(\frac{m_0}{\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)^{-\frac{1}{2}}) .

The first two `There was an error working with the wiki: Code[29]`

coefficients of the correction factor f(x) = (1 - x)^{-\frac{1}{2}} are:

: f(0) = (1 - 0)^{-\frac{1}{2}} = 1

: f'(0) = \frac{1}{2}(1 - 0)^{-\frac{3}{2}} = \frac{1}{2} .

So we approximate f(x) \approx f(0) + f'(0)x :

: E = c^2(m_0 (1 - x)^{-1/2}) \approx c^2 (m_0 (1 + \frac{1}{2}x)) = c^2 (m_0 (1 + \frac{1}{2} v^2/c^2 )) = m_0 c^2 + \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, and can be found by extending the expansion into a Taylor series by one more term:

: E \approx c^2 (m_0 (1 + \frac{1}{2} v^2/c^2 + \frac{3}{8} v^4/c^4 )) = m_0 c^2 + \frac{1}{2} m_0 v^2 + \frac{3}{8} m_0 v^4/c^2 .

For example, for a speed of 10 km/s the correction to the Newtonian kinetic energy is 0.07 J/kg (on a Newtonian kinetic energy of 50 MJ/kg) and for a speed of 100 km/s it is 710 J/kg (on a Newtonian kinetic energy of 5 GJ/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(\frac{1}{\sqrt{1 - (v/c)^2}} - 1) .

The relation between kinetic energy and `There was an error working with the wiki: Code[30]`

is more complicated in this case, and is given by the equation:

:E_k = \sqrt{p^2c^2+m_0^2c^4}-m_0c^2.

This can also be expanded as a `There was an error working with the wiki: Code[31]`

, 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 `There was an error working with the wiki: Code[5]`

, the expectation value of the electron kinetic energy, \langle\hat{T}\rangle, for a system of electrons described by the `There was an error working with the wiki: Code[6]`

\vert\psi\rangle is a sum of 1-electron operator expectation values:

:\langle\hat{T}\rangle = -\frac{\hbar^2}{2 m_e}\bigg\langle\psi \bigg\vert \sum_{i=1}^N \nabla^2_i \bigg\vert \psi \bigg\rangle

where m_e is the mass of the electron and \nabla^2_i is the `There was an error working with the wiki: Code[32]`

operator acting upon the coordinates of the i'th electron and the summation runs over all electrons.

The `There was an error working with the wiki: Code[7]`

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 \rho({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[\rho] = \frac{1}{8} \int \frac{ \nabla \rho({r}) \cdot \nabla \rho({r}) }{ \rho({r}) } d^3r

where T[\rho] is known as the Weizsacker kinetic energy functional.

`There was an error working with the wiki: Code[33]`

`There was an error working with the wiki: Code[34]`

`There was an error working with the wiki: Code[8]`

Serway, Raymond A. Jewett, John W. (2004). Physics for Scientists and Engineers, 6th ed., Brooks/Cole. ISBN 0-534-40842-7.

Tipler, Paul (2004). Physics for Scientists and Engineers: Mechanics, Oscillations and Waves, Thermodynamics, 5th ed., W. H. Freeman. ISBN 0-7167-0809-4.

Tipler, Paul Llewellyn, Ralph (2002). Modern Physics, 4th ed., W. H. Freeman. ISBN 0-7167-4345-0.

School of Mathematics and Statistics, University of St Andrews (2000). Biography of Gaspard-Gustave de Coriolis (1792-1843). Retrieved on 2006-03-03.

`There was an error working with the wiki: Code[1]`

, Wikipedia: The Free Encyclopedia. Wikimedia Foundation.

Comments