Intuitive physical concept for squared velocity in the context of energy Kinetic energy  $E_{kin} =  \frac{1}{ 2} m v^2 $ may be expressed in $kg \frac{ m^2 }{sec^2}$.The formula includes squared velocity. However, instead of squared velocity = v • v  it seems to be easier to imagine acceleration times distance = a • s as shown by the formula for work:
W = F • s = m • a • s
That means :


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*Energy is a constant force F applied over a distance s, force being mass times acceleration.


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*The unit $\frac{ m^2 }{sec^2}$ is divided into $\frac{ m}{sec^2}$ (acceleration) and $m$ (distance) .



I am looking for an intuitive model for the fact that energy includes squared velocity and where a • s is replaced by v • v.
Does there exist such an intuitive model in any domain of physics?
 A: It is not matter of units, or easier. The issue that you ask implies the energy conservation law:
if you invest mechanical work you obtain an increase of energy.
If the effect of the force is that the velocity of an object increases, then the mechanical work was spent for increasing the kinetic energy of the object.
So, imagine a force $F$ pushing an object of mass $m$ along a distance $s$. The mechanical work done by the force is
$\int_{s_0}^{s_1} F ds = m\int_{s_0}^{s_1} a \ ds$
Now, let's do a trick:
$m\int_{s_0}^{s_1} a \ ds = \int_{s_0}^{s_1} a \ dt \ \frac {ds}{dt} .$
But since $adt = dv$  and $ds/dt =v$ we get,
$\int_{s_0}^{s_1} F ds = m\int_{v_0}^{v_1} v \ dv = m\frac {v_1^2 - v_0^2}{2},$
where I changed the limits of integration in the integral in the middle because the variable of integration changed. Indeed the mechanical work in your problem increases the kinetic energy.
A: I think of V^2 as the rate of change of the surface area of an expanding sphere , it's radius changing at constant V , but the surface area A = Pi r^2 expanding at v^2. Does that help ?
Also in:  E = MC^2  
E = The total energy of a body at rest, V = 0
C^2 is a constant of proportionality, a real number, but not dimentionless - otherwise the formula would not be dimenionaly consistant. But this is beyond me!
