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 :

    1. Energy is a constant force F applied over a distance s, force being mass times acceleration.
    1. 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?

  • 3
    $\begingroup$ The kinetic energy formula is the result of an integration of $dW=Fds$: en.wikipedia.org/wiki/Kinetic_energy#Derivation $\endgroup$
    – CuriousOne
    Dec 21, 2014 at 8:28
  • $\begingroup$ Related: physics.stackexchange.com/q/535/2451 and links therein. $\endgroup$
    – Qmechanic
    Dec 21, 2014 at 9:18
  • $\begingroup$ I thik it verges on the impossibile to think intuitively about nonlinear relationships, that is why we define the energy in terms of force and distance and not in velocity squared. I think the human brain is not made for nonlinear relationships. One of the coolest examples is: guess how many ancestors you had that lived fivehundret years ago, then estimate it in a calculation, you will be surprised, because you just can't think nonlinearly on an intuitive level. But i might be wrong.^^ $\endgroup$
    – Kuhlambo
    Feb 14, 2015 at 14:13

2 Answers 2


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.

  • $\begingroup$ You might have proved the obvious equivalence of a • s and v • v, but I am looking for an intuitive description of energy using twice velocity instead of acceleration + distance. $\endgroup$
    – Moonraker
    Dec 21, 2014 at 10:35
  • $\begingroup$ Let me ask another way. Is it clear that applying a force to an object, its velocity would increase? And if yes, is your question why does it increase as $v^2$ and not simply $v$ is proportional to $F$ ? $\endgroup$
    – Sofia
    Dec 21, 2014 at 16:33
  • $\begingroup$ Energy is proportional to squared velocity, rest energy is proportional to mass and function of squared speed of light - how can we imagine this? There seems to be no intuitive approach. Example: the damage caused by a forest fire in a day may be proportional to its squared speed. This is intuitionally comprehensible, in the same way as that work W is proportional to acceleration a and to distance s. $\endgroup$
    – Moonraker
    Dec 21, 2014 at 17:05

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!

  • $\begingroup$ Your first idea is very interesting and may be felt as the first half of the answer. But is there a concept of energy which could use this squared velocity (in connection with a mass factor)? $\endgroup$
    – Moonraker
    Feb 14, 2015 at 14:02

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