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

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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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,

$W_{net}=\Delta K$

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  • $\begingroup$ 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. $\endgroup$ – StephenG Jun 28 at 11:20
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    $\begingroup$ I din't know it initially that it was duplicate. Though I will take care from next time. $\endgroup$ – Aditya Jain Jun 28 at 11:30
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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 $v$.

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