The formulae for kinetic energy and Joules

Click For Summary

Discussion Overview

The discussion revolves around the formulas for kinetic energy and joules, exploring the differences between them, the significance of the factor of 1/2 in the kinetic energy equation, and the definitions of energy and work in the context of physics. The scope includes conceptual clarification and technical reasoning.

Discussion Character

  • Conceptual clarification
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants express confusion regarding the factor of 1/2 in the kinetic energy formula and its absence in the definition of a joule.
  • Others clarify that joules are a unit of energy, not limited to kinetic energy, and emphasize the importance of distinguishing between physical quantities and units.
  • A participant states that the correct formula for kinetic energy is T = 1/2 mv^2 and explains the dimensional analysis leading to the definition of a joule.
  • Some participants argue that the factor of 1/2 is dimensionless and does not affect the unit equation for joules.
  • One participant discusses the relationship between work done and kinetic energy, mentioning the average speed of a decelerating mass and how it relates to the factor of 1/2.
  • Another participant elaborates on the SI unit definitions and the rationale behind avoiding conversion factors in derived units.
  • There are mentions of the complexities introduced by certain conversion factors in the SI system, particularly in electrodynamics.

Areas of Agreement / Disagreement

Participants express varying levels of understanding regarding the relationship between kinetic energy and joules, with some clarifying points while others remain confused. The discussion reflects multiple competing views and does not reach a consensus.

Contextual Notes

Some participants highlight the importance of correctly distinguishing between units and quantities, and there are unresolved questions about the implications of the factor of 1/2 in the context of energy definitions.

jamiebean
Messages
55
Reaction score
3
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 don't 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.
 
Physics news on Phys.org
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?
 
jamiebean said:
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 don't 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.
 
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).
 
vanhees71 said:
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 explanation.But i still don't understand why does the 1/2 gets eliminated when it comes to the equation of joule.
 
Dale said:
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 don't understand why does the 1/2 got eliminated in the equation of joule.
 
lomidrevo said:
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
 
jamiebean said:
i don't 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
 
jamiebean said:
i don't 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.
 
  • #10
jamiebean said:
i don't 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?
 
  • Like
Likes   Reactions: Dale
  • #11
Ibix said:
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.
 
  • Like
Likes   Reactions: osilmag, russ_watters, Dale and 1 other person
  • #12
jamiebean said:
i don't 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:
  • Like
Likes   Reactions: vanhees71
  • #13
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.
 
  • Like
Likes   Reactions: Dale

Similar threads

  • · Replies 138 ·
5
Replies
138
Views
9K
  • · Replies 41 ·
2
Replies
41
Views
4K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 15 ·
Replies
15
Views
3K
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 34 ·
2
Replies
34
Views
5K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K