= Kinetic energy
{title2=$K=\frac12mv^2$}
{wiki}
<Kinetic energy> is the <energy> associated with motion. For a nonrelativistic <point mass> of <mass> $m>0$, measured in a specified <inertial frame>,
$$
K=\frac12m\mathbf v\cdot\mathbf v=\frac12mv^2,
\qquad v=|\mathbf v|.
$$
Here $\mathbf v$ is <velocity> and $v$ is <speed>; $T$ is another common symbol for <kinetic energy>. The <dot product> makes $K$ nonnegative regardless of the direction of <velocity>. At fixed <mass>, doubling <speed> quadruples <kinetic energy>: a $2\,\mathrm{kg}$ <point mass> at $2\,\mathrm{m/s}$ has $4\,\mathrm J$, whereas at $4\,\mathrm{m/s}$ it has $16\,\mathrm J$. The <SI> units are $\mathrm{kg\,m^2/s^2}=\mathrm J$. Different <reference frames> can assign different <velocities> and therefore different <kinetic energies>. https://openstax.org/books/university-physics-volume-1/pages/7-2-kinetic-energy[OpenStax University Physics 1, §7.2: Kinetic energy]{ref}
A short outline explains why <kinetic energy> has this formula. In a nonrelativistic <inertial frame>, constant <mass> and <Newton's second law> give, using the <chain rule>,
$$
\frac{dK}{dt}=m\mathbf v\cdot\frac{d\mathbf v}{dt}
=\mathbf F_{\mathrm{net}}\cdot\mathbf v.
$$
Integrating over <time> proves the <work-energy theorem>, $W_{\mathrm{net}}=K_f-K_i$. The net <force> is the sum of all <forces>, and $W_{\mathrm{net}}$ is their total <work>, not merely the <work> of a chosen driving <force>. This route to final <speed> is useful when the <forces> as <functions> of <position> are known but the full motion as a function of <time> would be harder to find. https://openstax.org/books/university-physics-volume-1/pages/7-3-work-energy-theorem[OpenStax University Physics 1, §7.3: Work–energy theorem]{ref}
In the running example, a $2\,\mathrm{kg}$ <point mass> starts at rest and receives a constant net $3\,\mathrm N$ <force> through $4\,\mathrm m$. Its net <work> is $12\,\mathrm J$, so
$$
K_f=12\,\mathrm J,
\qquad v_f=\sqrt{\frac{2K_f}{m}}
=\sqrt{12}\,\mathrm{m/s}\approx3.4641\,\mathrm{m/s}.
$$
At any intermediate distance $x\ge0$, $K(x)=(3\,\mathrm N)x$ and $v(x)=\sqrt{2Fx/m}$: <kinetic energy> is a <linear function> of distance, while <speed> is proportional to the <square root> of distance. If a net opposing $3\,\mathrm N$ braking <force> subsequently acts for $4\,\mathrm m$, it does $-12\,\mathrm J$ of <work> and brings this <point mass> to rest. Negative net <work> decreases <kinetic energy>; it does not make <kinetic energy> negative. Beyond the stopping point the motion must be reconsidered, since the <point mass> cannot keep moving forward with a negative $K$.
\Image[/kinetic-energy-work-and-power.png]
{title=<Kinetic energy>, <velocity>, and <power> for a constant net <force>}
{description=A $2\,\mathrm{kg}$ <point mass> starts at rest under a net $3\,\mathrm N$ <force>. The left two panels show rightward <velocity> $v(x)=\sqrt{2Fx/m}$ and <kinetic energy> $K(x)=Fx$ over $0\le x\le4\,\mathrm m$, with separate axes for their different units. The right panel shows <instantaneous power> $P(t)=Fat$ up to $t_f\approx2.309401\,\mathrm s$ and <average power> $\overline P\approx5.19615\,\mathrm W$. The shaded area is $12\,\mathrm J$ of <work>. All values follow from the stated idealized model.}
<Kinetic energy> also makes the exchange with <potential energy> intuitive: a falling body's loss of <gravitational potential energy> can appear as increased <kinetic energy>. The formula above applies when <speed> is small compared with the <speed of light>; at higher <speeds>, <special relativity> changes the expression. https://www.feynmanlectures.caltech.edu/I_04.html[Feynman Lectures I, chapter 4, §§4–2 and 4–3: Energy of position and motion]{ref}
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