$$\frac{mv^2}{2}= Kinetic Energy$$ Can you explain me? What is purpose of $v^2$, $mv^2$, I am trying to understand the formula.


3 Answers 3


When people often ask 'how can I understand this equation or process?' what they really mean is whether they can break it down in terms of more familiar processes which they take for granted as basic. You can't keep on asking what or why or how eventually you have to reach a stage where no further simplification is possible and you have to take things as it is. Usually a good simplification is one that can explain the most or encompass the largest amount of information about the behaviour of the physical world. Newton's equation with his theory of gravity explains a very large portion of our daily experiences.

Let's say someone asks 'what is mass?'. In school you learn 'its the amount of substance' and you are happy with it. But what is substance? Let us say we say mass is the measure of Inertia. Then what is inertia? Inertia is the property due to which different objects are accelerated to different velocities under the same force. I now have to define force or at least a notion of 'same' force. Say a spring kept elongated from its equilibrium position by 1cm. What I have done then is just create a measure of my intuitive idea of what I call force. May be a cruder definition is a push by my friend Jack. If you go down this road you will either go in 'definition' circles or indefinitely define new terms.

Of course as you learn more and more physics these definitions change. In quantum mechanics there is no force. So it makes no sense to talk about motion or acceleration to define mass as they are not fundamental physical properties in our theory. What you need to understand then is ultimately physical quantities are objects that helps us understand patterns in nature. Usually you turn them into numbers that fit in some mathematical theory that represents some behaviour of our observed universe.

Now having said that, the best way to understand how that specific equation of kinetic energy comes is simply in terms of conservation. How do we know something is conserved? We find it from experiments. It was found from experiments that in fact this quantity, historically $\sum_i m_iv_i^2$, remained conserved. Say we do some collision experiments. We see from these experiments that in some collision processes $\sum_i mv^2$ (how the half came is another story) is conserved. We can also see from these experiments that the quantity $mv$ remains a constant(if we have some a priory measure of mass) in all cases. As our experimental knowledge of the world increased we found out that this quantity is more fundamental than we previously thought and is part of a more fundamental principle of Energy conservation.

Now if one does define work done on an object by the application of a force as

\begin{eqnarray} W = \int_{x_1}^{x_2}F(x)\,\mathrm d x.\\ =\int_{x_1}^{x_2}m \frac{dv}{dt} dx\\ =\int_{x_1}^{x_2}m \frac{dx}{dt} dv\\ =\int_{v_1}^{v_2}m v dv\\ =\frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2 \end{eqnarray} Which means any work done on the body manifests as a change in $\frac{1}{2}mv^2$ of the object. Which gives us some idea into what the quantity is. But nevertheless the equation itself as far as I know doesn't hold any more meaning than this or can be 'understood' in better terms.

I don't think I have done justice to your question and I feel that my answer is more like a rant. But If you are still perplexed, its always nice to look up some history to see how the founding fathers of physics where equally unaware and how they made sense of it. Our problem simply is that we take some things for granted that are not at all very obvious.


Can you clarify a bit what is it what you are trying to understand? The $v^2$ term tells you that the kinetic energy of an object with mass $m$ increases with the square of the velocity. $$KE\propto v^2$$ So if the objects doubles its velocity $v_{f}\rightarrow 2\cdot v_{i}$ this means that its kinetic energy increases by a factor of $4$.

$$KE_{i}\propto v_{i}^2$$ $$KE_{f}\propto v_{f}^2$$ But $v_{f}=2\cdot v_{i}$ which means that: $$KE_{f}\propto \left(2\cdot v_{i}\right)^2=4\cdot v_{i}^2$$ And this means that the kinetic energy increased $4$ times: $$\frac{KE_{f}}{KE_{i}}=\frac{4\cdot v_{i}^{2}}{v_{i}^{2}}=4$$

$m$ is just the mass of the object. The kinetic energy of an object proportional to its mass. Let's consider two object with different masses. The mass of object $1$ is $m_{1}$ and the mass of object $2$ is $m_{2}$, where $m_{1}\neq m_{2}$. Let's say $m_{1}>m_{2}$. We can write the kinetic energy expression for both objects (ignoring the $\tfrac{1}{2}$ term):

$$KE_{1}\propto m_{1}v_{1}^{2}$$ $$KE_{2}\propto m_{2}v_{2}^{2}$$

If both objects have the same velocity $v=v_{1}=v_{2}$, we can write:

$$KE_{1}\propto m_{1}v^{2}$$ $$KE_{2}\propto m_{2}v^{2}$$

Dividing $\tfrac{KE_{1}}{KE_{2}}$ we get:


So if $m_{1}>m_{2}$, that means that $KE_{1}>KE_{2}$.

The $\tfrac{1}{2}$ term is just a constant, and it is irrelevant from the physics point of view. It is just a number that you have to multiply the expression with and it results from the mathematical calculation. Also, this kinetic energy expression is valid only when $\tfrac{v}{c}\ll 1$. The general expression for the kinetic energy is:


which means the difference between the total energy of the particle and its rest energy.

Expanding further:

$$KE=mc^2\left(\gamma -1\right)$$

$$KE=mc^2\left(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} -1\right)$$

We can take this term $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ and write its Taylor expansion:


In our case $f(x)$ is:

$$\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=\left(\frac{1}{1-\frac{v^2}{c^2}}\right)^{1/2}=\left(1-\frac{v^2}{c^2}\right)^{-1/2}$$ We are interested in $\tfrac{v}{c}\ll 1$. $$\left(1-\frac{v^2}{c^2}\right)^{-1/2}=1+\frac{1}{2}\frac{v^2}{c^2}+\frac{3}{8}\left(\frac{v^2}{c^2}\right)^2+\cdots$$

Going back and inserting this expansion into the general kinetic energy formula:

$$KE=mc^2\left[\underbrace{\left(1+\frac{1}{2}\frac{v^2}{c^2}+\frac{3}{8}\left(\frac{v^2}{c^2}\right)^2+\cdots\right)}_{\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}} -1\right]$$


So we can consider that:

$$KE\approx \frac{1}{2}mv^2 \quad, \text{ for $\tfrac{v}{c}\ll 1$}$$


Well, $\frac{mv^2}{2}$ is the result which we get from work energy theorem.$$dW_{net}=\vec{F_{net}}.d\vec{s}=m\vec{a_{net}}.d\vec{s}$$ After solving this equation further,we get:-

$$W_{total}=\Delta K.E=\frac{m(v_{f}^2-v_{i}^2)}{2}$$

Hope this helps!


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