Is kinetic energy just as arbitrary as potential energy? Potential energy $U$ can be defined up to an arbitrary additive constant $c$ because $$F=-\dfrac{d(U+c)}{dx}=-\dfrac{dU}{dx}=ma$$ And therefore the equation of motion remains unchanged. I think the same holds for kinetic energy $T$ using a similar reasoning and I want to make sure that I'm getting it right. 
In a system where conservation of mechanical energy holds true $$T+U=\dfrac{1}{2}mv^2+U=\text{constant}$$  differentiating with respect to position $x$ we get $$\dfrac{dT}{dx}=-\dfrac{dU}{dx}=F=ma$$ therefore by the same token one expects that kinetic energy can be defined up to an arbitrary additive constant such that $$T_c=\dfrac{1}{2}mv^2+c$$ where usually we prefer to set $c=0$ for simplicity.
Is this fact right about kinetic energy?
 A: That's right but be careful about that constant that takes an interesting value from special relativity. Anyhow, I prefer this other approach. Consider
$$
m\frac{d{\bf v}}{dt}={\bf F}
$$
and multiply both members of this equation by ${\bf v}$. You will get
$$
m{\bf v}\cdot\frac{d{\bf v}}{dt}={\bf F}\cdot{\bf v}
$$
that means
$$
\frac{d}{dt}\left(\frac{1}{2}mv^2\right)={\bf F}\cdot{\bf v}.
$$
You can integrate in time obtaining
$$
\frac{1}{2}mv^2+c=\int{\bf F}\cdot{\bf v}dt.
$$
The constant is generally fixed by the problem at hand, e.g. by computing the work of the force on the right hand side.
A: My answer would be yes; in the same frame of reference the work done on the mass $m$ will be the same for every observer:
$$W_{1-2}=\int_{t_1}^{t_2}\vec{F}.\vec{v}dt=
\int_{t_1}^{t_2}\frac{d}{dt}\left(\frac{mv^2}{2}\right)dt=\frac{mv_2^2}{2} - \frac{mv_1^2}{2} = \Delta E_{kin}$$
so therfore we can a time-independent constant to the function in the second integral  - $\frac{mv^2}{2}$ -  without changing the result. And because kinetic energy can be definied as this function, the kinetic energy is definied up to a constant.
