Possible Duplicate:
Why is there a $\frac 1 2$ in $\frac 1 2 mv^2$?

Hèllo, I have a question about kinetic energy formula.

As you know, in kinetic energy formula, we have:


Okay. And we know, Joule (energy unit), is: $\large J= kg~(\frac{m}{s})^2$

(Guys, please light me up if I'm wrong.)

Here's my question:

Why do we have a $\frac{1}{2}$ in our formula? Why do we divide our $mv^2$?

Please answer in simple word.

  • 1
    $\begingroup$ Possible duplicate: physics.stackexchange.com/q/27847/2451 $\endgroup$
    – Qmechanic
    Sep 9, 2012 at 11:39
  • $\begingroup$ wow! I understood nothing from that page:( I'm 1st high-school student. Can anyone write a simple answer, please? $\endgroup$ Sep 9, 2012 at 11:48
  • $\begingroup$ the derivation of the formula is complicated, and i think the simplest explanation requires at least calculus and linear algebra. en.wikipedia.org/wiki/Kinetic_energy#Derivation (or you could try reading newton's principia, but good luck.) that's why the formula exists -- so that you can just plug variables into it without having to worry about deriving it! personally, i think of it as similar to the difference between a compiled program and program source code. $\endgroup$
    – ixtmixilix
    Sep 9, 2012 at 12:15
  • $\begingroup$ Another dupicate (more general though): physics.stackexchange.com/questions/71119/… $\endgroup$ Jul 15, 2013 at 4:23

5 Answers 5


The factor $\frac12$ comes in because we're integrating the equation $$ \frac{\mathrm dE}{\mathrm dv}=mv $$ once.

Less abstract and only using basic arithmetics, the story goes like this:

When accelerating a body by applying a (constant) force $F$ along a distance $\Delta s$, the body gains energy according to $$ \Delta E=F\Delta s $$ which is just the definition of (mechanical) work.

According to Newton's second law $F=ma$. We also have $\Delta s \approx v\Delta t$ and thus $$ \Delta E\approx mav\Delta t $$ This relationship is only approximate because during any finite time interval $\Delta t$, the value $v$ changes as the whole point of the exercise was accelerating the body.

Now, as $a\Delta t=\Delta v$ we have $$ \Delta E \approx mv\Delta v $$

But where does the factor $\frac12$ come in? From basic calculus: $$ \Delta(v^2)=(v+\Delta v)^2-v^2=2v\Delta v+(\Delta v)^2\approx 2v\Delta v $$ which yields $$ \Delta E\approx\frac12m\Delta(v^2)=\Delta(\frac12 mv^2) $$ and thus $$ E\approx\frac12 mv^2 + \mathrm{const} $$ If we go from finite to infinitesimal time intervals, the equations become exact and we no longer need to assume a constant force.

A short introduction to differential calculus as relevant to this particular example:

At time $t = t_0$ the body has a velocity $v(t_0)=v_0$. After a time $\Delta t$, the body has the velocity $v(t_0+\Delta t)=v_0 + \Delta v$.

The value of $v^2$ at time $t=t_0$ is of course $v^2(t_0)=v(t_0)^2=v_0{}^2$. What's the value of $v^2$ at time $t=t_0+\Delta t$? $$ v^2(t_0 + \Delta t)=v(t_0+\Delta t)^2 = (v_0+\Delta v)^2 $$ On the other hand, we also have $$ v^2(t_0 + \Delta t) = v^2(t_0) + \Delta(v^2) = v_0{}^2 + \Delta(v^2) $$ and thus $$ \begin{align*} \Delta(v^2) &= (v_0 + \Delta v)^2 -v_0^2 \\ &= v_0^2 + 2v_0\Delta v + (\Delta v)^2 - v_0{}^2 \\ &= 2v_0\Delta v + (\Delta v)^2 \end{align*} $$ We're interested in the instantaneous values, ie the change as we take the limit $\Delta t \rightarrow 0$. This means that $\Delta v$ becomes arbitrarily small as well and we're in particular able to ignore higher powers like $(\Delta v)^2$ and get $$ \Delta(v^2)\approx 2v_0\Delta v $$ or $$ \frac{\Delta(v^2)}{\Delta v}\approx 2v_0 $$ This procedure is so useful that it got its own formalism and symbolic notation $$ \frac{\mathrm{d}(v^2)}{\mathrm{d}v}=2v $$ after taking the limit $\Delta v\rightarrow 0$.

  • 2
    $\begingroup$ Thanks, mister. More or less, I got it. But the only part of your answer that I didn't understand is that part you wrote about calculus. (No wonder, We haven't studied calculus yet). However if do I temporarily accept that Δ(v^2)=2vΔv, everything is right. You'll be a good teacher! I think other guys were gonna tell this to me, but you answered very simple-to-understand. $\endgroup$ Sep 9, 2012 at 16:29
  • $\begingroup$ @MoctavaFarzán: hope my additions help clear things up; stackexchange is of course no substitute for taking a course on calculus, though... $\endgroup$
    – Christoph
    Sep 9, 2012 at 19:01
  • $\begingroup$ Khan Academy is a great resource for learning about calculus. $\endgroup$
    – jcora
    Sep 14, 2012 at 21:57

So, you require a simple answer... Let us consider a body of mass $m$ at rest. Initial velocity $u=0$. Now, The body moves with a velocity $v$.

Force on the accelerating body is $F=ma$, $$\implies F=m\frac{dv}{dt}$$

The small amount of work done in moving the body over a small distance $ds$ is $$dw=F.ds$$ $$dw=m\frac{ds}{dt}dv=mvdv$$ Hence, the total work done in accelerating the body from $0$ to $v$ is $$W=\int_0^v dw=\int_0^v mvdv$$ $$\implies W={mv^2 \over 2}$$ This work done is stored as the kinetic energy of moving body... $K.E=\frac{1}{2}mv^2$

In this method, Half comes through integration. There is no other explanation easier than this, I think so... But, the actual derivation is provided by Wikipedia


You'll notice, as you study physics more, that a factor of one-half often accompanies squared quantities. For example:




These are kinetic energy, capacitor energy and inductor energy respectively.

If you've studied calculus based physics, you know that the time rate of change of kinetic energy is power and that power is the product of force and velocity.

$\frac{d}{dt}KE = f \cdot v = ma \cdot v = m \frac{dv}{dt} \cdot v$

Integrating both sides with respect to $t$ yields:

$KE = \int mv\ \frac{dv}{dt}dt = \int mv\ dv = \frac{1}{2}mv^2$

If you're not familiar with calculus yet, the above will not make much sense but, as you become more familiar with it, you'll recognize the factor of one-half often arises because of an integration.

  • $\begingroup$ that's what helped me understand the formula: the mantra, it's been integrated, it's been integrated, it's been integrated, every time i look at a body in movement. $\endgroup$
    – ixtmixilix
    Sep 9, 2012 at 12:23
  • 1
    $\begingroup$ @AlfredCentauri: You could also include the rotational kinetic energy $E=\frac{1}{2}I\omega^2$ $\endgroup$ Sep 9, 2012 at 12:43
  • $\begingroup$ As a(n) (unfortunate) counterexample, $A=\pi r^2$. The pattern would be followed if we write $A=\frac{1}{2}(2\pi)r^2$. Obviously $2\pi$ should have been the natural constant, and not $\pi$. A shame... $\endgroup$
    – QuantumDot
    Sep 9, 2012 at 14:56
  • $\begingroup$ @QuantumDot $A=\frac{1}{2}\tau r^2$ math-blog.com/2010/06/28/forget-pi-here-comes-tau $\endgroup$
    – Greg
    Jul 5, 2013 at 14:34

Here is a small line of arguments which is not a DERIVATION, but hopefully it will give you intuitive picture between factor of 2 and definition of different energies in physics.

As you know energy is conserved so if the planet orbits the sun its potential energy is constantly converted to kinetic and vice versa.

Our question is how to represent mechanical Energy through the notion of speed?

Lets assume following example: We release a test body from the hight H and let it free fall to the groung.

At the beginning the body has the potential energy $E=m g H$, we know that when the body hits the ground it will have potential energy equal to 0 (since $H=0$) and it means all its initial potential energy will be converted to kinetic energy.

So we have $K= E=m g H$.

Now all we need to do is to represent this energy through the speed. We can do it by noticing that for a free falling body $v=g t$ then $t=\sqrt{\frac{2 H}{g}}$ since $H=\frac{g t^2}{2}$ and finally $g H=\frac{v^2}{2}$.

So for intial energy we have $E=K= mgH= m \frac{v^2}{2}$

Here factor of 2 came from the definition of distance traveled by body under constant acceleration of g.

Hope it helps your intuition.


Although I agree with the many answers here as to the origin of the $\frac{1}{2}$ factor, there is an important point that has not been made yet.

The fact is, the factor of 1/2 is only there because of the system of units used to measure mass. We could easily change our system of units in such a way that would make the kinetic energy just $mv^2$.

Actually, being comfortable with changing from one system of units to another is an important and useful trick. For example, this is frequently done to have a system of units where the speed of light is defined as equal to 1, which is nice because it means you don't have to write $c$ all over the place.

  • 3
    $\begingroup$ Well, if do we change units, we can make it a little easier to calculate $K=mv^2$ but in other hand, we should change all the other energy formulas like this: $U=2mgh$. Or I misunderstood you? $\endgroup$ Sep 9, 2012 at 16:41
  • $\begingroup$ What you say about changing other formulas is correct. The important physics is that momentum is proportional to velocity, and energy is proportional velocity squared. The constants of proportionality are related by a factor of 2, but the choice of system of units is arbitrary. $\endgroup$
    – user442
    Sep 12, 2012 at 3:53
  • $\begingroup$ @Matt: as Moctava pointed out, if all the other energy formulas need to be changed to have a factor of 2, some one would ask the meaning of 2, instead of 1/2 in this question. So we are basically facing the same question regardless of the unit of choice. $\endgroup$
    – Codism
    May 30, 2013 at 20:22
  • $\begingroup$ This is not strictly true. The equation $T = \frac{1}{2}mv^2$ is always true. If you perform a change of units to obtain a relation of the form $T = kmv^2$ where $k \neq \frac{1}{2}$, the constant $k$ will be dimensional. The units of $k$ must then also appear in the equation. $\endgroup$
    – 13509
    May 25, 2020 at 20:08
  • $\begingroup$ IMHO mass isn't responsible for the 1/2. V is responsible. The problem with V is that it's variable, and slows down to impart energy and momentum. The 1/2 shows up in the calculus. Compare with c which doesn't slow down. c is always the same speed. There is no 1/2 factor in: E=mc^2. $\endgroup$ Nov 7, 2023 at 15:05

Not the answer you're looking for? Browse other questions tagged or ask your own question.