# The formulae for kinetic energy and Joules

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## Main Question or Discussion Point

The formula for kinetic energy is 0.5x kg x v^2, where v= velocity
The formula for joules is kg x v^2

I dont understand why does the difference exist
and what is the difference between joules and kinetic energy
isnt joule the unit for energy??

Thank you guys for helping me.

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Dale
Mentor
Are you confused about the factor of 1/2 in the expression for KE or about the factor of 1 for the definition of a Joule?

The formula for kinetic energy is 0.5x kg x v^2, where v= velocity
The formula for joules is kg x v^2

I dont understand why does the difference exist
and what is the difference between joules and kinetic energy
isnt joule the unit for energy??

Thank you guys for helping me.
As you said, joule is a unit of energy (btw. not only kinetic). The problem is that both equations in your post are wrong. In both of them you mix symbols for physical quantities (velocity) and symbols for units (kg). Have you searched a bit on google for correct equations? Even Wikipedia does a good job for the basic topics like this one.

vanhees71
Gold Member
2019 Award
You also should clearly distinguish units and quantities. In your case the right formula is
$$T=\frac{1}{2} m v^2.$$
Each quantity has a dimension within the used system of units. In the SI units the dimensions are $[m]=\text{kg}$ and $[v]=\text{m}/\text{s}$. Consequently the dimension for energy is $[T]=\text{kg} \, \text{m}^2/\text{s}^2=1 \, \text{J}$. Thus the unit Joule is a derived unit, just used as a shortcut and to honour Joule, who did a lot of important work to establish the concept of energy (particularly in thermodynamics).

You also should clearly distinguish units and quantities. In your case the right formula is
$$T=\frac{1}{2} m v^2.$$
Each quantity has a dimension within the used system of units. In the SI units the dimensions are $[m]=\text{kg}$ and $[v]=\text{m}/\text{s}$. Consequently the dimension for energy is $[T]=\text{kg} \, \text{m}^2/\text{s}^2=1 \, \text{J}$. Thus the unit Joule is a derived unit, just used as a shortcut and to honour Joule, who did a lot of important work to establish the concept of energy (particularly in thermodynamics).
Thanks for your explaination.But i still dont understand why does the 1/2 gets eliminated when it comes to the equation of joule.

Are you confused about the factor of 1/2 in the expression for KE or about the factor of 1 for the definition of a Joule?
i dont understand why does the 1/2 got eliminated in the equation of joule.

As you said, joule is a unit of energy (btw. not only kinetic). The problem is that both equations in your post are wrong. In both of them you mix symbols for physical quantities (velocity) and symbols for units (kg). Have you searched a bit on google for correct equations? Even Wikipedia does a good job for the basic topics like this one.
sorry..i mixed up units and quantities.thats why the equation went wrong

i dont understand why does the 1/2 got eliminated in the equation of joule.
1/2 is dimensionless factor, appearing in the equation for kinetic energy. Once you do a dimensional analysis of this equation, you get what you are calling here "equation of joule". The dimensionless factor can be fully ignored in this analysis, it has no influence on the "composition" of Joule. I suggest you to read this article: http://hyperphysics.phy-astr.gsu.edu/hbase/units.html

A.T.
i dont understand why does the 1/2 got eliminated in the equation of joule.
Because it has no units, so it changes nothing in the unit equation.

Ibix
i dont understand why does the 1/2 got eliminated in the equation of joule.
They are definitions of different things, in short. One Joule is the amount of work done by a force of one Newton moving through a distance of one metre. On the other hand, $mv^2/2$ is the amount of energy possessed by a mass of 1kg moving at 1m/s. Why would you expect these quantities to be the same?

jbriggs444
Homework Helper
2019 Award
They are definitions of different things, in short. One Joule is the amount of work done by a force of one Newton moving through a distance of one metre. On the other hand, $mv^2/2$ is the amount of energy possessed by a mass of 1kg moving at 1m/s. Why would you expect these quantities to be the same?
In this particular case, one can trace the factor of two to a notion of averaging (or to the area of a triangle).

If you have a 1kg mass decellerating under an applied force of 1 Newton, it will come to rest after an interval of one second. During that second, its average speed will be 1/2 meter per second. (It started at 1 meter per second, ended at 0 meters per second and changed speed at a uniform rate). Thus, it will have covered a total of 1/2 meter during that time and performed 1/2 Joule of work. E=1/2mv^2.

The triangle comes in if you integrate force over distance and look at the area under the curve. 1/2 base times height. Or integrate Fv dv to get 1/2 Fv^2.

Dale
Mentor
i dont understand why does the 1/2 got eliminated in the equation of joule.
In the SI units the derived units are always defined as straight combinations of the base unit. The purpose is to avoid any need for conversion factors. So the design of the SI system requires that J = 1 kg m^2/s^2 and N = 1 kg m/s^2 and W = 1 kg m^2/s^3. That is simply one of the rules that the SI committee has agreed to follow in the definition of all the derived units.

Now, suppose that the SI committee had decided to allow such conversion factors and suppose they had chosen to define the conversion factor
k = 2 J/(kg m^2/s^2)

Then KE = k m v^2 which is not really any simpler, and PE = k m g h which is less simple, and W = k F d which is less simple, and so forth. It would not simplify the KE equation, and it would make all other energy equations more complicated. Plus everyone would have to remember the conversion factor, which nobody likes.

Last edited:
vanhees71
Well, sometimes the SI introduces pretty annoying conversion factors :-). The worst are $\epsilon_0$ and $\mu_0$ in electrodynamics. Of course they do this for good reasons to make the units simpler to use in everyday live. Theoretical physicists can always convert between natural units, i.e. Heaviside Lorentz units in e&m and SI units. For even more natural units, where$\epsilon_0$ and $\mu_0$ and thus also $c$ are all set to one HL and SI are identical.