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Line 1: − + − K ≡ [[mass|m]][[velocity|v]]<sup>2</sup> / 2 for a point mass and <math> K = {1 \over 2}mV^2 + {1 \over 2} I \omega ^2 </math> for a rigid body, where I is the body's moment of inertia and omega is the body's angular velocity.+ + + + + + + + + + + + + + + + + + + + + + + + − Σ''W'' = ΔK = mv<sub>f</sub><sup>2</sup> / 2 - mv<sub>i</sub><sup>2</sup> / 2+ + + + + + + + + + + + + + + + + + + + + + + + − Where v<sub>i</sub> is [[speed]] at t = 0 and v<sub>f</sub> is speed at [[time]] = t.+ − Kinetic energy is a scalar and has the same units as work (i.e. [[Joule]]). + − Note that if the mass of an object is increased, the increase in kinetic energy increases linearly; if the velocity of an object is increased, the increase in kinetic energy increases quadratically. For example, doubling the mass of an object doubles its kinetic energy; doubling its velocity quadruples its kinetic energy. +
Restructured, converted Formulae to math Formulae and added derivation
'''Kinetic energy''' represents the [[energy]] associated with the [[motion]] of an object.<ref>Serway and Beichner, ''Physics for Scientists and Engineers'', Fifth Edition</ref> It is defined as:
'''Kinetic energy''' represents the [[energy]] associated with the [[motion]] of an object.<ref>Serway and Beichner, ''Physics for Scientists and Engineers'', Fifth Edition</ref> It is defined as the work done by a force to accelerate that object from rest to some speed <math> v </math>, in the absence of any other [[force|forces]] acting upon the object. Kinetic energy is a scalar and has the same units as work (i.e. [[Joule]]).
==Classical mechanics==
===Translational kinetic energy===
In [[classical mechanics]], the translational kinetic energy of a ridid object, <math> K </math>, can be found as:
<math> K = \frac{1}{2} m v^2</math>
Where
:<math> m </math> is the [[mass]] of the object
:<math> v </math> is the [[velocity]] of the object
===Rotational kinetic energy===
The rotational kinetic energy of a rigid object is:
<math> K = {1 \over 2} I \omega ^2 </math>
Where
:<math> I </math> is the [[moment of inertia]] of the object
:<math> \omega </math> is the angular velocity of the object
===Work-Energy theorem===
The change of kinetic energy is equal to the total [[work]] done on it by the resultant of all [[force]]s acting on it. For a point mass this can be expressed as:
The change of kinetic energy is equal to the total [[work]] done on it by the resultant of all [[force]]s acting on it. For a point mass this can be expressed as:
<math> \Sigma W = \Delta K = \frac{1}{2} m v_{f}^{2} - \frac{1}{2} m v_{I}^{2} </math>
Where
:<math> v_i </math> is the initial [[speed]]
:<math> v_f </math> is the final [[speed]]
Note that if the mass of an object is increased, the increase in kinetic energy increases linearly; if the [[velocity]] of an object is increased, the increase in kinetic energy increases [[quadratic equation|quadratically]]. For example, doubling the mass of an object doubles its kinetic energy; doubling its velocity quadruples its kinetic energy.
===Derivation of translational kinetic energy===
The [[work]] done by a force accelerating an object from rest, which is the kinetic energy is:
<math> W = K = \int F dx</math>
From Newton's second law, the [[force]], <math> F</math>, is <math> F =\frac{dp}{dt} </math>. Hence we can make the substitution and use the chain rule
<math> K = \int \frac{dp}{dt} dx = \int \frac{dp}{dx} \frac{dx}{dt} dx </math>
This is the same as
<math> K = \int v dp </math>
In [[classical mechanics]], [[momentum]] is given by <math> p = mv </math>. Differentiating and substituting into the above equation results in
<math> K = \int^{u}_{0} m v dv </math>
We want to integrate between 0 and the speed of the object, <math> u </math> as this defines kinetic energy. Performing the integration reveals that the kinetic energy is, as expected, the following:
<math> K = \frac{1}{2} mv^2 </math>
== Kinetic Energy in Relativity ==
== Kinetic Energy in Relativity ==
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