# How to understand the process of the kinetic energy formula? [duplicate]

$$\frac{1}{2} mv^2 = K.E.$$ What is the purpose of $$v^2$$; why can't use $$v$$ instead?

Why this question is arising because of that I learned the whole process of speed= distance/time. (I know How it works) Velocity = displacement/time. Acceleration, Average speed, This whole formula have some connection, that I learned or realised by questioning the formula. But I can't understand $$\frac{1}{2} mv^2$$. Without memorizing I want to understand each part of the formula.

## marked as duplicate by Alfred Centauri, StephenG, Thomas Fritsch, Bill N, knzhouJun 28 at 12:43

To simply put, Kinetic energy can be calculated by the basic process of computing the work (W) that is done by a force (F). If the body has a mass of m that was pushed for a distance of d on a surface with a force that’s parallel to it.

$$W=F.d=m.a.d$$

The acceleration in this equation can be substituted by the initial $$(v_i)$$ and final $$(v_f)$$ velocity and the distance. This we get from the kinematic equations of motion.

$$W=m.a.d\\ \\ =m.d.\frac{v_{f}^{2}-v_{i}^{2}}{2d}\\ \\ =m.\frac{v_{f}^{2}-v_{i}^{2}}{2d}\\ \\ =\frac{1}{2}.m.v_{f}^{2}-\frac{1}{2}.m.v_{i}^{2}$$

The Kinetic Energy’s (K) basic quantity $$\frac{1}{2}mv^{2}$$ changes when a particular sum of work is acted upon an object.

$$K.E=\frac{1}{2}mv^{2}$$

The total work that is done on a system is equivalent to the change in kinetic energy. Thus,

$$W_{net}=\Delta K$$

• As a general rule please do not add new answers when duplicate (or almost duplicate) questions exist. Also note that sometimes these questions fall under the homework-type question policy and insufficient prior research (including e.g. Wikipedia and a search for prior answers) by OPs can be a reason for closing. – StephenG Jun 28 at 11:20
• I din't know it initially that it was duplicate. Though I will take care from next time. – Aditya Jain Jun 28 at 11:30

The answer to this question has the roots in the symmetry properties of space and time. I will choose the lagrangian formalism of classical mechanics to answer the question, where the lagrangian $$\mathcal{L}$$ is a scalar function which is the difference between kinetic energy and potential energy.

We know that space is homogeneous and isotropic, and time is homogeneous. For a free particle, it follows that the lagrangian $$\mathcal{L}$$ should have the following properties:

1. $$\mathcal{L}$$ should not depend on the position coordinate.
2. $$\mathcal{L}$$ should not depend on the velocity vector. Rather it should depend on the magnitude of the velocity, i.e., some power of the velocity vector.
3. $$\mathcal{L}$$ should not depend on the time coordinate.

So the general form of the lagrangian would be $$\mathcal{L}(x,v,t)=\alpha v^n$$ where $$\alpha$$ is a constant. Now, we can evaluate the momentum by using the relation $$p=\frac{\partial\mathcal{L}}{\partial v}=\alpha nv^{n-1}$$ We know that the momentum is a linear function of the velocity. This is possible only when $$n=2$$ in the above expression.

The lagrangian function is written as $$\mathcal{L}=T-U$$, where $$T$$ is the kinetic energy and $$U$$ is the potential energy. Since we are considering a free particle (which has only kinetic energy), the lagrangian (choosing $$n=2$$) is $$\mathcal{L}=T=\alpha v^2$$ Thus, the kinetic energy is proportional to $$v^2$$ and not any other power of $$v$$.