Editing Kinetic energy (section)

===Kinetic energy of rigid bodies===
In [[classical mechanics]], the kinetic energy of a ''point object'' (an object so small that its mass can be assumed to exist at one point), or a non-rotating [[rigid body]] depends on the [[mass]] of the body as well as its [[speed]]. The kinetic energy is equal to half the [[Multiplication|product]] of the mass and the square of the speed. In formula form:

<math display="block">E_\text{k} = \frac{1}{2} mv^2</math>

where <math>m</math> is the mass and <math>v</math> is the speed (magnitude of the velocity) of the body. In [[SI]] units, mass is measured in [[kilogram]]s, speed in [[metres per second]], and the resulting kinetic energy is in [[joule]]s.

For example, one would calculate the kinetic energy of an 80&nbsp;kg mass (about 180&nbsp;lbs) traveling at 18&nbsp;metres per second (about 40&nbsp;mph, or 65&nbsp;km/h) as

<math display="block">E_\text{k} = \frac{1}{2} \cdot 80 \,\text{kg} \cdot \left(18 \,\text{m/s}\right)^2 = 12,960 \,\text{J} = 12.96 \,\text{kJ}</math>

When a person throws a ball, the person does [[work (physics)|work]] on it to give it speed as it leaves the hand. The moving ball can then hit something and push it, doing work on what it hits. The kinetic energy of a moving object is equal to the work required to bring it from rest to that speed, or the work the object can do while being brought to rest: '''net force × displacement = kinetic energy''', i.e.,

<math display="block">Fs = \frac{1}{2} mv^2</math>

Since the kinetic energy increases with the square of the speed, an object doubling its speed has four times as much kinetic energy. For example, a car traveling twice as fast as another requires four times as much distance to stop, assuming a constant braking force. As a consequence of this quadrupling, it takes four times the work to double the speed.

The kinetic energy of an object is related to its [[momentum]] by the equation:
<math display="block">E_\text{k} = \frac{p^2}{2m}</math>

where:
*<math>p</math> is momentum
*<math>m</math> is mass of the body

For the ''translational kinetic energy,'' that is the kinetic energy associated with [[rectilinear motion]], of a [[rigid body]] with constant [[mass]] <math>m</math>, whose [[center of mass]] is moving in a straight line with speed <math>v</math>, as seen above is equal to

<math display="block"> E_\text{t} = \frac{1}{2} mv^2 </math>

where:
*<math>m</math> is the mass of the body
*<math>v</math> is the speed of the [[center of mass]] of the body.

The kinetic energy of any entity depends on the reference frame in which it is measured. However, the total energy of an isolated system, i.e. one in which energy can neither enter nor leave, does not change over time in the reference frame in which it is measured. Thus, the chemical energy converted to kinetic energy by a rocket engine is divided differently between the rocket ship and its exhaust stream depending upon the chosen reference frame. This is called the [[Oberth effect]]. But the total energy of the system, including kinetic energy, fuel chemical energy, heat, etc., is conserved over time, regardless of the choice of reference frame. Different observers moving with different reference frames would however disagree on the value of this conserved energy.

The kinetic energy of such systems depends on the choice of reference frame: the reference frame that gives the minimum value of that energy is the [[center of momentum]] frame, i.e. the reference frame in which the total momentum of the system is zero. This minimum kinetic energy contributes to the [[invariant mass]] of the system as a whole.

====Derivation====
=====Without vector calculus=====
The work W done by a force ''F'' on an object over a distance ''s'' parallel to ''F'' equals 

<math display="block">W = F \cdot s.</math>

Using [[Newton's second law]] 

<math display="block">F = m a</math> 

with ''m'' the mass and ''a'' the [[Acceleration#Uniform_acceleration|acceleration]] of the object and 

<math display="block">s = \frac{a t^2}{2}</math>

the distance traveled by the accelerated object in time ''t'', we find with <math>v = a t</math> for the velocity ''v'' of the object

<math display="block">W = m a \frac{a t^2}{2} = \frac{m (at)^2}{2} = \frac{m v^2}{2}.</math>

=====With vector calculus=====
The work done in accelerating a particle with mass ''m'' during the infinitesimal time interval ''dt'' is given by the dot product of ''force'' '''F''' and the infinitesimal ''displacement'' ''d'''''x'''

<math display="block">\mathbf{F} \cdot d \mathbf{x} = \mathbf{F} \cdot \mathbf{v} d t = \frac{d \mathbf{p}}{d t} \cdot \mathbf{v} d t = \mathbf{v} \cdot d \mathbf{p} = \mathbf{v} \cdot d (m \mathbf{v})\,,</math>

where we have assumed the relationship '''p'''&nbsp;=&nbsp;''m''&nbsp;'''v''' and the validity of [[Newton's second law]]. (However, also see the special relativistic derivation [[#Relativistic kinetic energy of rigid bodies|below]].)

Applying the [[product rule]] we see that:

<math display="block">d(\mathbf{v} \cdot \mathbf{v}) = (d \mathbf{v}) \cdot \mathbf{v} + \mathbf{v} \cdot (d \mathbf{v}) =  2(\mathbf{v} \cdot d\mathbf{v}).</math>

Therefore (assuming constant mass so that ''dm'' = 0), we have

<math display="block">\mathbf{v} \cdot d (m \mathbf{v}) = \frac{m}{2} d (\mathbf{v} \cdot \mathbf{v}) = \frac{m}{2} d v^2  = d \left(\frac{m v^2}{2}\right).</math>

Since this is a [[total differential]] (that is, it only depends on the final state, not how the particle got there), we can integrate it and call the result kinetic energy:

<math display="block">E_\text{k} = \int_{v_1}^{v_2}\mathbf{p}d\mathbf{v} =  \int_{v_1}^{v_2}m\mathbf{v}d\mathbf{v} = {mv^2\over 2}\bigg\vert_{v_1}^{v_2} = {1\over 2}m(v_2^2-v_1^2).</math>

This equation states that the kinetic energy (''E''<sub>k</sub>) is equal to the [[integral]] of the [[dot product]] of the [[momentum]] ('''p''') of a body and the [[infinitesimal]] change of the [[velocity]] ('''v''') of the body. It is assumed that the body starts with no kinetic energy when it is at rest (motionless).