How does most energy get transfered to mass at high relavistic speeds, but mostly to movement at low speeds? From what I understand about relativity, it is impossible to accelerate a massive object to the speed of light because it's mass would become infinite. Once an object is moving close to the speed of light, adding energy to it will result (mostly) in increasing its mass, not it's speed. However, at low speeds, adding the same amount of energy will result an in a significant increase in speed. How does this work physically? Where is the energy going when it increases the object's mass? How come at low speeds most of the energy gets transfered to kinetic energy, but at high speeds most of the energy gets transferred to it's mass?
 A: The formula for kinetic energy that applies for a body at all possible speeds (not just speeds much less than the speed of light, $c$) is
$$\text{KE}=mc^2\left(\frac 1{\sqrt{1-v^2/c^2}}-1\right)$$
$m$ is a constant for the body: its mass (formerly called 'rest mass' and denoted by $m_0$).
Physicists no longer use the concept of 'relativistic mass', a mass that varies with speed. [Calling $\tfrac m {\sqrt{1-v^2/c^2}}$ 'relativistic mass' is now acknowledged to be unhelpful, suggesting as it does that when a body moves faster something else happens to it as well as it simply moving faster! Or just note that 'relativistic mass' falls foul of Occam's razor.
What's the alternative? $c^2\left(\frac 1{\sqrt{1-v^2/c^2}}-1\right)$ is best thought of as a kinematic factor that, SR demands, must replace its low-speed limit, $\tfrac 1 2 v^2$, in the Newtonian formula, $\text {KE}=\tfrac 1 2 m v^2$.
So as the body gets faster it gains kinetic energy. The energy gained for a given small change of speed increases dramatically as $v$ approaches $c$. Equivalently, the increase in speed for a given extra amount of KE acquired becomes smaller and smaller as $v$ approaches $c$. This is dictated by the kinematic factor noted earlier. The energy transfer is from whatever is pushing the body to the body's kinetic energy. It's as simple as that, I'm afraid!
I said earlier that $m$ was a constant for the body. But there are ways of changing a body's mass $m$. Increasing the body's speed is not one of them. But making the body participate in an inelastic collision in which its internal energy, $U$, changes is (even if the mass change is too small to measure for a macroscopic body)! Here Special Relativity has forced a radical concept shift ($\Delta m=(1/c^2)\Delta U)$ in dynamics. No concept shift is demanded for kinetic energy – just a different formula!
A: I think that the formula you are thinking about when asking this question is
$$ E=mc^2=\gamma m_0 c^2,$$
where $E$ is the total energy of the object, $m$ it's relativistic mass, $m_0$ it's rest mass and $\gamma=1/(\sqrt{1-(v/c)^2})$ is the Lorentz factor with $c$ the speed of light. You see, in this formula we make no distinction between mass and kinetic energy.
If we now add energy to our object, its mass and velocity increase at the same time and energy is not 'split' between a kinetic and a mass term. To see how this looks like at low energies, we can take a Taylor expansion in $E$ for $(v/c)<<1$:
$$E \approx m_0 c^2+\frac{1}{2}m_0v^2\,.$$
So at low energies, it looks like mass does not increase with velocity and instead of $(E,v,m)$ increasing at the same time, it is only $(E,v)$. Thus, it is useful to introduce the concept of kinetic energy, $E_{kin}=\frac{1}{2}m_0v^2$. So as a conclusion, the 'splitting' of energy you are talking about is just an artefact of the low-energy description.
