I am having doubts regarding why $1/2$ is present in kinetic energy and if this is conventional why cant we say potential energy is $2mgh$ and kinetic energy $mv^2$. So is this $1/2$ conventional or not, and how?

In early 18th century kinetic energy or vis viva was considered proportional to $mv^2$ and by 1829 it had been properly defined as in its modern form of $mv^2/2$ in coriolis's book " Calcul de l'Effet des Machines " and in the same book was mechanical work introduced .As I am unable to read the above mentioned paper, would anybody please mention how did Coriolis derive notions of work and kinetic energy?

In addition can someone please elaborate Ron Maimon's answer, especially what does he mean by energy mixing with momentum?

  • $\begingroup$ The factor of 1/2 follows from a trivial integration. There is no choice about it if we define dW=Fds, which is a fairly natural definition. $\endgroup$ – CuriousOne Feb 22 '16 at 7:59
  • $\begingroup$ @CuriousOne How can dW=Fds be a natural definition ? how would anybody relate force to energy anyhow without invoking energy first ? is it just only because work and energy have the same dimensions ? $\endgroup$ – Faiz Iqbal Feb 22 '16 at 8:09
  • $\begingroup$ @FaizIqbal Your coriolis link points me to an article by Smeaton on wind and water mills? $\endgroup$ – Farcher Feb 22 '16 at 8:20
  • $\begingroup$ You don't relate force to energy but to work. Is it twice as hard to push a cart that is twice as heavy? You formalize that with simple machines en.wikipedia.org/wiki/Simple_machine and that's pretty much that as far as "natural-ness" is concerned. Kinetic energy trivially follows from that. We are teaching all of this in high school physics, by the way, but I don't think that it becomes all that obvious for students why we are teaching it. The progression of terms is logical... in hindsight, not necessarily while one is learning it. $\endgroup$ – CuriousOne Feb 22 '16 at 8:21
  • $\begingroup$ @farcher i have edited it , actually it is a complete book $\endgroup$ – Faiz Iqbal Feb 22 '16 at 9:04

Simple enough,

Just consider Newton's second law of dynamics:

$$ \frac{d\mathbf{p}}{dt} = \sum_i \mathbf{{F_i}} = \mathbf{F} $$

where bold characters represent vectors and $\mathbf{F}$ is just the total resulting force.

In order to give a definition of the kinetic energy, let's just consider an ordinary system that will enable us to build some intuition. Here we'll consider an object of constant mass $m$ and velocity $\mathbf{v}$. Then by definition of the momentum $\mathbf{p}$:

$$ \mathbf{p} = m \mathbf{v}. $$

Now let's insert this in Newton's second law and perform a dot product with $\mathbf{v}$ on both sides with the idea to latter integrate the formula :

$$ \begin{eqnarray} \mathbf{v} \cdot \frac{d (m ~\mathbf{v})}{dt} &=& \mathbf{v} \cdot \mathbf{F}\\ \frac{d }{dt} \left(\frac{m ~\mathbf{v}^2}{2}\right) &=& \mathbf{v} \cdot \mathbf{F}\\ d \left(\frac{m ~\mathbf{v}^2}{2}\right) &=& \mathbf{v} \cdot \mathbf{F} ~ dt \end{eqnarray} $$

Then we can already see that the natural quantity $K = \frac{m ~\mathbf{v}^2}{2}$ varies if the quantity $ \mathbf{v} \cdot \mathbf{F} ~ dt$ varies. We then can decide to call this new quantity $K$ the kinetic energy since it is a new quantity that depends on the speed only and we can refine the expression of $ \mathbf{v} \cdot \mathbf{F} ~ dt$ :

$$ \frac{d\mathbf{s}}{dt} \cdot \mathbf{F} ~ dt = \mathbf{F} \cdot d\mathbf{s} $$

where we used that $\mathbf{v} = \frac{d\mathbf{s}}{dt}$.

As we did for $K$ we'll name this new quantity $W$. We'll call it work since it represents an amount of force carried along some path. Work properly defined is the integral of this quantity between two points A and B.

$$ W_{AB} = \int_A^B \mathbf{F} \cdot d\mathbf{s} $$

So this should settle the matter.

One last point, this little demonstration was meant to show how the $\frac{1}{2}$ factor naturally arrises from the fundamental equations of dynamics. However, nothing forced us except some good sense, to chose $K = \frac{m ~\mathbf{v}^2}{2}$ as the definition of the kinetic energy and it has, now, been accepted as an international convention or, let's just say, usage. Therefore, anyone giving the amount of kinetic energy would use this relation for the kinetic energy -- otherwise, this person would have to specify it before hand, and faces a cumbersome process that may loose this person's audience.

Remark : In the derivation, we didn't use the condition that the mass was constant, it never left the differential operator $\frac{d}{dt}$.

  • $\begingroup$ what difference would have had it made if the candition of a constant mass was made (in reference to remark) $\endgroup$ – Faiz Iqbal Feb 23 '16 at 4:18
  • $\begingroup$ None, it would just have reduced the generality of the demonstration. $\endgroup$ – A.G. Feb 23 '16 at 9:18
  • $\begingroup$ your whole post comes down to the fact that work is a definition and not a theory and now i am quite familiar with that . fine ! so what do you say about ron maimon's answer mentioned in my question ? $\endgroup$ – Faiz Iqbal Feb 23 '16 at 9:23
  • $\begingroup$ Yes it is a definition, like your name is a definition that you carry with you every day. If you are looking for theories, then they will refer to those definitions. Work and kinetic energy naturally arise from physical equations and it makes a lot of sense to stick with $\frac{1}{2}mv^2$, that is what we are both saying, Ron Maimon and I. So Ron answered your question as well. $\endgroup$ – A.G. Feb 23 '16 at 12:15
  • $\begingroup$ thanks could you also please elaborate Ron Memons concept as an answer here $\endgroup$ – Faiz Iqbal Feb 23 '16 at 12:18

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