Talk:Kinetic energy: Difference between revisions

Mathematics B‑class High‑priority

This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.MathematicsWikipedia:WikiProject MathematicsTemplate:WikiProject Mathematicsmathematics articles B This article has been rated as B-class on Wikipedia's content assessment scale. High This article has been rated as High-priority on the project's priority scale.

Physics B‑class Top‑importance

This article is within the scope of WikiProject Physics, a collaborative effort to improve the coverage of Physics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.PhysicsWikipedia:WikiProject PhysicsTemplate:WikiProject Physicsphysics articles B This article has been rated as B-class on Wikipedia's content assessment scale. Top This article has been rated as Top-importance on the project's importance scale.

Energy B‑class Top‑importance

This article is within the scope of WikiProject Energy, a collaborative effort to improve the coverage of Energy on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.EnergyWikipedia:WikiProject EnergyTemplate:WikiProject Energyenergy articles B This article has been rated as B-class on Wikipedia's content assessment scale. Top This article has been rated as Top-importance on the project's importance scale.

Template:WP1.0

Archives

• Archive 1

Expositions of Kinetic Energy (Newtonian) are more effective in close association to work and potential energy - a simplified version of Hamiltonian Mechanics.

The absence of such material seems to have caused a great deal of confusion in the discussion page Blablablob (talk) 05:32, 17 September 2008 (UTC)[reply]

in the phrase "This is done by binomial approximation. Indeed, taking Taylor expansion for square root and keeping first two terms we get:"

I found that I had to take three terms of the Taylor series, not two.

can someone confirm?

-gary —Preceding unsigned comment added by 212.179.48.132 (talk) 13:42, 25 September 2008 (UTC)[reply]

Suppose one is trying to get approximations to

${\displaystyle E={\frac {mc^{2}}{\sqrt {1-(v/c)^{2}}}}\,.}$

The essential part is

${\displaystyle (1-\beta ^{2})^{-1/2}\,}$

where ${\displaystyle \beta =v/c\,.}$ If the Taylor series is expanded using ${\displaystyle x=\beta \,}$ as the independent variable in which you are expanding, then three terms are required to get

${\displaystyle {\frac {d(1-x^{2})^{-1/2}}{dx}}=-{\frac {1}{2}}(1-x^{2})^{-3/2}\cdot -2x=x(1-x^{2})^{-3/2}\,}$

${\displaystyle {\frac {d^{2}(1-x^{2})^{-1/2}}{{dx}^{2}}}=(1-x^{2})^{-3/2}+3x^{2}(1-x^{2})^{-5/2}\,}$

${\displaystyle (1-\beta ^{2})^{-1/2}\approx 1+0\cdot \beta +{\frac {1}{2}}\cdot \beta ^{2}\,.}$

However, if you use ${\displaystyle x=\beta ^{2}\,}$ as the variable in which you are expanding, then only two terms are needed to get

${\displaystyle {\frac {d(1-x)^{-1/2}}{dx}}=-{\frac {1}{2}}(1-x)^{-3/2}\cdot -1={\frac {1}{2}}(1-x)^{-3/2}\,}$

${\displaystyle (1-\beta ^{2})^{-1/2}\approx 1+{\frac {1}{2}}\cdot \beta ^{2}\,.}$

So the two different ways of expanding give different lengths. OK? JRSpriggs (talk) 17:46, 25 September 2008 (UTC)[reply]

Yes, I think that's fine. Thank you for taking the time to explain. Gary —Preceding unsigned comment added by 84.108.230.143 (talk) 06:51, 26 September 2008 (UTC)[reply]

The idea that the amount of kinetic energy in a body "depends on the relationship between that object and the observer" is a difficult one, and the article currently doesn't come close to explaining why and how this should be so in a way that an educated layman can understand. I'd like to see a much clearer and more accessible explanation of this, and I'd encourage those who understand the subject to address this issue. —Preceding unsigned comment added by 86.161.43.41 (talk) 23:42, 15 January 2009 (UTC)[reply]

Yes, this paragraph was a mess. I cleaned it up and removed that "sea also below"-thing and the "gravitational"-remark. I.m.o. This is not the place for this. DVdm (talk) 18:41, 16 January 2009 (UTC)[reply]

As a non-mathematician, I once had a teacher walk me through this concept. I recall he pointed out the COMBINED kinetic energy of the earth and a one-kilogram weight impacting, and the concept of the relative speeds. I recall the moment of clarity when I did the math. Unfortunately, this is lost in the dim recesses of my mind! A similar example should be appreciated by many students. Mydogtrouble (talk) 15:19, 24 August 2009 (UTC)[reply]

Is it possible to be clearer that the mass referred to is rest mass? A '0' subscript maybe? —Preceding unsigned comment added by 194.94.232.72 (talk) 12:19, 28 August 2009 (UTC)[reply]

The 0-subscript is a bit old-fashioned. I made a little change, moving the reference to rest mass up a bit. DVdm (talk) 13:41, 28 August 2009 (UTC)[reply]

Correct me if I'm wrong, but isn't it true that the 1/2 in (1/2)mv^2 is a proportionality constant that only applies to the SI system of units? If you wanted to express kinetic energy in the british engineering system you would use (1.29e-3)mv^2 = KE in BTU's. This article makes it seem as if the formula for KE is derived when in fact it is an empirical formula. The only thing we know regardless of units is that KE is proportional to mv^2. I think it's important to try to make it clear where that 1/2 comes from. I got most of this information from the external link at the bottom of the article on the conservation of energy.--70.185.112.226 (talk) 22:33, 15 February 2009 (UTC)[reply]

There are other kinds of energy than kinetic, for example thermal energy, and work energy. You can choose to define energy with any proportionality constant you like, but once you define it in any one area, you're now stuck in all other others. Thus, if you define heat energy as C*ΔT, for example (3/2)nRΔT for ideal gases, or work energy as Force*Distance (both of which quantities are equal to each other) then that forces you to define THAT energy as 1/2 mV^2 in the same units. SBHarris 22:51, 15 February 2009 (UTC)[reply]

Something is bothering me. Now Leibniz knows well that mgh=1/2mv^2. Why does he insist KE is mv^2. To me his aim is not to derive an equation that describe how object falls but rather the philosophy of KE. —Preceding unsigned comment added by 68.11.142.236 (talk) 15:08, 17 February 2009 (UTC)[reply]

• Perhaps so. But with Count Rumford figuring out that F*D = Q in mechanical work, and Joule verifying that plus also that mgh = Q (I think he did that on his honeymoon at some Falls), we can either call this "vis-viva" mv^2 and say then that it's twice the heat, 2Q, or we can bite the bullet and say it's the same as Q, and thus 1/2 mv^2. Clearly the important relationship, is that the quantity in motion that scales with heat goes as v^2, not v. After you know that, you can chose your scale factor, but if you chose it as 1/2, you can simply say it's all the same thing; the motion equivalent of heat and work. Instead of needing two names and a conversion factor of 2 between them, which seems silly. SBHarris 02:09, 18 February 2009 (UTC)[reply]

In Lp space the L^p norm of x is given by

${\displaystyle \ \|x\|_{p}=\left(|x_{1}|^{p}+|x_{2}|^{p}+\cdots +|x_{n}|^{p}\right)^{1/p}}$

(so the L² norm is the familiar Euclidean norm (see Pythagorean theorem), while the distance in the L¹ norm is known as the Manhattan distance).
AFAIK, in L^p space KE=kv^p and PE=F^(p-1) * D
just-emery (talk) 23:58, 22 June 2009 (UTC)[reply]

The mathematics is well-known to us. But the real world (where "kinetic energy" has a meaning) has no relationship to L^p for p≠2. You cannot just make up stuff and put it in an article. JRSpriggs (talk) 09:27, 23 June 2009 (UTC)[reply]

I didnt say that the real world wasnt L2. I am simply pointing out that the reason the equations take the form they do in the real world is because the real world is L2. This may be well known to you but its certainly not common knowledge. I cant imagine why you would think I was 'making stuff up' but of course I'm sure that you acted in good faith so you must have just misunderstood me. just-emery (talk) 16:20, 23 June 2009 (UTC)[reply]

Yes, the real world is L² (in the 3 known spatial dimensions, at least).

If you say that "the reason the equations take the form they do in the real world is because the real world is L2.", then you are also saying implicitly that the equations would have a different form in L^p for p≠2 (indeed, you specified the supposed form that they would have). However, you provided no evidence that kinetic energy would have such a form, or even that the notion of kinetic energy could be defined at all in such a hypothetical world. JRSpriggs (talk) 20:26, 7 July 2009 (UTC)[reply]

The derivation of the formula ${\displaystyle E_{k}={\frac {1}{2}}mv^{2}}$ that I was shown seems much simpler than the one currently in the article. I'm wondering what others think. It is as follows:

${\displaystyle W=F\times x\,}$

${\displaystyle E_{k}=\int F\,dx}$

${\displaystyle E_{k}=\int ma\,dx}$

Assuming constant mass then

${\displaystyle E_{k}=m\int a\,dx}$

${\displaystyle E_{k}=m\int {\frac {dv}{dt}}\,dx}$

${\displaystyle E_{k}=m\int {\frac {dx}{dt}}\,dv}$

${\displaystyle E_{k}=m\int v\,dv}$

${\displaystyle E_{k}={\frac {1}{2}}mv^{2}}$

A few steps could probably be omitted, I just wanted to show each step explicitly. Any thoughts or corrections would be appreciated. Ginogrz (talk) 03:05, 7 July 2009 (UTC)[reply]

Hi Ginogrz. I like your alternative derivation. It is simpler and clearer. One major objection I have about the present derivation is that it is not supported by any in-line citations to allow independent verification, as is required by WP:Verifiability. I encourage you to substitute your derivation, but before you do so please establish a suitable source. When you add your derivation be sure to include at least one in-line citation to show where your information comes from. Information on citing sources is available at WP:CITE. Dolphin51 (talk) 06:17, 7 July 2009 (UTC)[reply]

One reason for the current derivation is that the first few steps can also be used as part of the relativistic derivation, which is not possible with your version. This helps to make clear that the key change is in the definition of linear momentum.

Also notice that historically, Newton actually defined force as the time derivative of linear momentum, not as mass times acceleration. See the second law at Newton's laws. JRSpriggs (talk) 20:14, 7 July 2009 (UTC)[reply]

Ah, I can see how that is beneficial for setting up the reader for the relativistic derivation. However, it seems like it might be better to explain the ${\displaystyle E_{k}=\int \mathbf {v} \cdot d\mathbf {p} }$ statement in the relativistic section rather than the Newtonian derivation since kinetic energy is a pretty fundamental concept in beginning physics courses, which often do not require a knowledge of dot product properties or vector calculus. Also the current derivation doesn't seem to flow well with the preceding section in which the formula is introduced (a fairly uncomplicated section), being that it is in a totally different realm of knowledge. This is only my opinion though, any other thoughts would be nice.

I found a source for the derivation I suggested in a textbook, however since the section centers around Newton, it would only make sense to use his historical definition of force like you said. You could then just simply change the derviation to be

${\displaystyle E_{k}=\int F\,dx}$

${\displaystyle E_{k}=\int {\frac {d(mv)}{dt}}\,dx}$

Constant mass

${\displaystyle E_{k}=m\int {\frac {dv}{dt}}\,dx}$

${\displaystyle E_{k}=m\int {\frac {dx}{dt}}\,dv}$

${\displaystyle E_{k}=m\int v\,dv}$

${\displaystyle E_{k}={\frac {1}{2}}mv^{2}}$

Although I suppose I would have to find a new source for this one. Ginogrz (talk) 03:02, 8 July 2009 (UTC)[reply]

I prefer the approach suggested by Ginogrz, rather than the one suggested by JRSpriggs. The former approach is much more compatible with the principle of make technical articles accessible. Wikipedia articles should begin with the simplest information so that it is accessible to the greatest number of readers, and work towards the more complex information later in the article. Students and others are introduced to the concept of kinetic energy long before they need to assimilate the concept of scalar products, vector products and relativity. Dolphin51 (talk) 03:50, 8 July 2009 (UTC)[reply]

Perhaps you should add another step, otherwise it is not entirely obvious that

${\displaystyle m\int {\frac {dv}{dt}}\,dx=m\int {\frac {dv}{dt}}{\frac {dx}{dt}}\,dt=m\int {\frac {dx}{dt}}\,dv\,.}$ JRSpriggs (talk) 04:29, 9 July 2009 (UTC)[reply]

I don't know if that would help considering ${\displaystyle {\frac {dv}{dt}}\,dx={\frac {dx}{dt}}\,dv}$ is a pretty simple identity and adding those extra steps would only elongate the derivation. Ginogrz (talk) 06:38, 15 July 2009 (UTC)[reply]

Isn't according kinetic energy formula should be easier to drive bike on slow pedals rotation-fast wheel rotation mode instead fast pedals rotation-slow wheel rotation mode? How such example fits here? And why flying stone in kosmos have about square bigger kinetic energy than momentum? If object to object transmiting all momentum to eatch over, so how is transmitted kinetic energy? Supose 1 kg mass stone flying in cosmos at 100 m/s transmit his energy to don't moving relatively to this stone, to over bigger stone with mass 10 kg momentum. So smaller stone stop and bigger stone got speed 10 m/s. How supose here fit in kinetic energy? —Preceding unsigned comment added by Taupusensteinas (talk • contribs) 09:11, 15 October 2009 (UTC)[reply]

Regarding the bicycle question : see this. Bicycle stability physics is actually quite complex. With regard to your second question : kinetic energy is defined to be the integral of momentum with respect to changes in velocityUser A1 (talk) 10:09, 15 October 2009 (UTC)[reply]

To Taupusensteinas: I presume that you are not a native speaker of English since your grammar is so poor that it is hard to understand you.

If the collision of the stones is elastic, then the 1 kg stone cannot just stop when it hits the 10 kg stone (assumed to be initially at rest). If the collision is inelastic, then the lost kinetic energy becomes a mixture of: heat, sound, and damage to the structure of the stones (chemical energy). JRSpriggs (talk) 10:51, 15 October 2009 (UTC)[reply]

From article:

The work done accelerating a particle during the infinitesimal time interval dt is given by the dot product of force and displacement:

${\displaystyle \mathbf {F} \cdot d\mathbf {x} =\mathbf {F} \cdot \mathbf {v} dt={\frac {d\mathbf {p} }{dt}}\cdot \mathbf {v} dt=\mathbf {v} \cdot d\mathbf {p} =\mathbf {v} \cdot d(m\mathbf {v} ).}$

Applying the product rule we see that:

${\displaystyle d(\mathbf {v} \cdot \mathbf {v} )=(d\mathbf {v} )\cdot \mathbf {v} +\mathbf {v} \cdot (d\mathbf {v} )=2(\mathbf {v} \cdot d\mathbf {v} ).}$

Therefore (assuming constant mass), the following can be seen:

${\displaystyle \mathbf {v} \cdot d(m\mathbf {v} )={\frac {m}{2}}d(\mathbf {v} \cdot \mathbf {v} )={\frac {m}{2}}dv^{2}=d\left({\frac {mv^{2}}{2}}\right).}$

Since this is a total differential (that is, it only depends on the final state, not how the particle got there), we can integrate it and call the result kinetic energy:

${\displaystyle E_{k}=\int \mathbf {F} \cdot d\mathbf {x} =\int \mathbf {v} \cdot d\mathbf {p} ={\frac {mv^{2}}{2}}.}$

This equation states that the kinetic energy (E_k) is equal to the integral of the dot product of the velocity (v) of a body and the infinitesimal change of the body's momentum (p). It is assumed that the body starts with no kinetic energy when it is at rest (motionless).

As I see the force is itself particle with velocity v, so this derivations shows not kinetic energy, but what velocity will reach particle if he will fly with acceleration, so velocity will be quadratic per linear time and liner distance x. Let me try to explain. F=dp/dt, so F already is particle of mass m flying at velocity v in small time interval delta-t=dt. And now again we see v*dt, which means distance traveled per very small time. So here is multiplication of distance traveled at speed v with mass and with rising speed. Time also rising. If we take constant time say dt=10 seconds. And constant speed v=100 m/s. m=1 kg. And dv/dt is average speed. So this equation describes what will be if object of mass m=1 kg and flying at acceleration dv/dt=100/10=10 m/s/s will be multiplicated with distance, what he fly through 10 seconds at speed 100 m/s. This distance is 1000 meters. So this distance is muliplied by mass and acceleration. Acceleration times distance and mass and only 10000 kg remaining from all this trick. From formula should be taken dv. Acceleration may be imposible to write in math phorm and aliens etc don't using it in math like symbol (dv/dt). And it have tricky phorm m/s^2, which frozen in time.

${\displaystyle \mathbf {F} \cdot d\mathbf {x} =\mathbf {F} \cdot \mathbf {v} dt={\frac {m}{dt}}\cdot \mathbf {v} dt=\mathbf {v} \cdot m.}$

So m do not moving yet, but why not it can't be devided by dt. Mass m egzist in cosmos in whatever time frame. And it multiplicated with v gives us old good mv momentum. How object can have already some speed dv if it do not have? Or if object have already acceleration then he have already energy before he was multiplied by v and before goting kinetic energy. There somthing dirty with this derivation of kinetic energy and acceleration dimension.

But here good question: Who will move faster after collision, bed (of mass m=40 kg) in which metal ball of mass 1 kg stroke at 10 m/s in soft side or same mass as bed metal box say of mass m=40 kg? Ball almost do not reflecting from bed soft side, so isn't ball transmit bigger momentum and kinetic energy to soft bed than to metal box? Or this ball stoping springs force is converted to heat and like that? Who will got bigger velocity: bed or box after collision with ball? —Preceding unsigned comment added by Taupusensteinas (talk • contribs) 10:25, 17 October 2009 (UTC)[reply]

The center of mass moves with a speed of 10/41 m/s before and after collision, in both cases, so if the ball stays with the bed each of the two have this speed. If the ball bounces back the box will have a higher speed than 10/41 m/s. The kinetic energy is more in the second case, in the first more is converted to heat etc.--Patrick (talk) 07:49, 7 February 2010 (UTC)[reply]

I calculated this as an example of an object at relativistic speeds: 1 kg object at 0.8 c would have kinetic energy of about 60 PJ (accurately it would be 59917011915787842 and two thirds J). That would make the kinetic energy in such an object little less than a third of the yield of Tsar Bomba. Is this anywhere near the actual value? How would I calculate the "yield" of this kind of projectile? --Khokkanen (talk) 23:35, 7 June 2010 (UTC)[reply]

This Discussion page is primarily about the article Kinetic energy. If you don't receive a prompt answer to your question I suggest you delete it and re-post it at WP:Reference desk/Science. Dolphin (t) 02:20, 8 June 2010 (UTC)[reply]

${\displaystyle E_{k}={\frac {mc^{2}}{\sqrt {1-(v/c)^{2}}}}-mc^{2}}$. The factor gamma-1 at 0.8 c is 2/3. Thus an object will have a kinetic energy 2/3 that of mc^2. A kg is 21 megatons, so this energy is 2/3 of that, or 14 megatons. This is 14/50 = 28% of Tsar bomba, so your calculation is indeed correct (quite a glowbomb, as Vinge would say). Another way of looking at this is that the 50 megaton yeild of Tsar Bomba is 50/21 = 2.38 kilograms of heat, light, and other forms of active energy. SBHarris 02:52, 8 June 2010 (UTC)[reply]

{{editsemiprotected}}

Please consider changing "The kinetic energy of an object is the extra energy..." to "The kinetic energy of an object is the energy..."

Adding "extra" prior to energy is confusing and is not consistent with other accepted definitions of KE available. TSRibeye (talk) 07:22, 11 June 2010 (UTC)[reply]

No. I think that the word "extra" is necessary for correctness. Objects have energy even when they are not moving. I want to make it clear that the kinetic energy is distinct from that rest energy. JRSpriggs (talk) 07:42, 11 June 2010 (UTC)[reply]

While it is true and correct that objects do have energy when not moving, generally accepted and used definitions of KE do not include the word "extra". The definition of KE includes the phrase "..the energy of motion.." and potential energy. "Motion" removes any question as to KE referring to energy "at rest". At a minimum it may be helpful to remove "extra" from the initial definition and then discuss how this "extra" energy is different.

Spigotmap- I was following Wikipedia's directions for suggesting edits to protected content by using the editsemiprotected template. If I'm doing this the wrong way, please let me know how else to go about suggesting edits and correct the Wiki directions. (TSRibeye (talk) 21:31, 14 June 2010 (UTC))[reply]

 Not done for now: please establish a consensus for this alteration before using the {{Edit semi-protected}} template. SpigotMap 12:19, 11 June 2010 (UTC)[reply]

To TSRibeye: I do not see what your problem is with that word. "Extra" works with "due to its motion" to express an idea. Some people define "kinetic energy" in a way which includes rest energy. That definition conflicts with the one used in this article. "Extra" helps to draw attention to the distinction.

Do not feel that we have to slavishly follow some particular source. There are many thousands of sources out there which do things in a variety of different ways. It is more important to be internally consistent and clearly present the concepts. JRSpriggs (talk) 06:57, 15 June 2010 (UTC)[reply]

My problem with the word "extra" is that it does not add anything to the definition, and may be confusing to someone looking to understand the theory. As for "slavishly" following a particular source, I have yet to see "extra" used in any other definition. If you would like I can cite the resources I used, but it may be easier if you can cite your source for the definition. I feel it is more important to be consistent across the internet while clearly explaining the concept-neither of which apply to the use of "extra". All I'm asking is it be removed from the definition, but the inclusion of energy at rest be included in the next sentence. This will allow for a definition that is in line with the majority of main stream sources, and ensure the understanding of what the extra motion means. (TSRibeye (talk) 19:37, 15 June 2010 (UTC))[reply]

Sources for defining kinetic energy taken from the first two search pages of a google search for "kinetic energy". Note none of them include the word "extra" in the definition. Please cite your reference for "extra energy". TSRibeye (talk) 21:02, 24 June 2010 (UTC)[reply]

• www.physicsclassroom.com/class/energy/u5l1c.cfm
• hyperphysics.phy-astr.gsu.edu/hbase/ke.html
• www.schools.utah.gov/curr/science/sciber00/8th/forces/.../potkin.htm
• physics.about.com/od/energyworkpower/f/KineticEnergy.htm
• scienceworld.wolfram.com/physics/KineticEnergy.html
• wiki.answers.com/Q/What_is_kinetic_energy
• http://www.wisegeek.com/what-is-kinetic-energy.htm
• www.thefreedictionary.com/kinetic+energy

I agree with TSRibeye's latest theme. If authoritative sources can be found, and cited, for use of the word extra then use of that extra word in this article is reasonable. Conversely, if no authoritative source can be found then the word extra should not be used in the article. The argument put forward by JRSpriggs is a sound one and I don't disagree with it, but in the absence of a citable source it is merely original research. At WP:V there is the instruction that The threshhold for inclusion in Wikipedia is verifiability, not truth. Dolphin (t) 02:16, 25 June 2010 (UTC)[reply]

You must be joking. Now we have to have a source for the exact choice of wording, for grammar? Do not be ridiculous. JRSpriggs (talk) 07:09, 25 June 2010 (UTC)[reply]

You can try distracting our attention by use of expressions like You must be joking and Do not be ridiculous but I won't be distracted by such things. On Wikipedia we try to discuss things objectively. TSRibeye and I have made a very simple and totally reasonable request. So far you haven't acknowledged the legitimacy of that request, nor have you identified a source for your belief that kinetic energy is extra energy.

I have checked my Physics textbook and it doesn't suggest that kinetic energy is extra energy. It is at pains to point out that we are always interested in changes of energy rather than absolute values. For example, when I am sitting at my computer I can choose a reference frame that gives me a velocity of zero and a kinetic energy of zero. Alternatively I can use some other reference frame that gives me a non-zero velocity and a non-zero kinetic energy. If I stand up and walk away at 1 metre per second my velocity and kinetic energies are different in the two reference frames but the change of kinetic energy is equal to the work done in both cases. Similarly, whether my gravitational potential energy is measured relative to the floor, or ground level, or sea level, or the center of the Earth is unimportant because it is only the change in potential energy that conforms to the laws of physics. So if I choose a reference frame in which an object's velocity is zero I am entitled to say that object's kinetic energy is zero. If I choose to say the object's temperature is equal to my arbitrary datum I am entitled to say its internal energy is zero; and I can also say its energy is zero. Consequently I think it is reasonable to challenge your view that kinetic energy should be defined to be extra energy, and ask you to identify your source. Dolphin (t) 07:53, 25 June 2010 (UTC)[reply]

"Extra energy" DOES talk about only about a changes in energy. Kinetic energy is the "extra energy" due to relative motion of an object or center of mass of a system of objects, and this is frame-dependent. There is a sort of "minimum energy" in SR which occurs when the system has zero-momentum (center of mass frame). All this neglects gravity and curved space (mass in general relativity). But what's wrong with defining kinetic energy as the change in energy due to motion. Change = delta = "extra." They are basically all the same concept, no? SBHarris 19:14, 25 June 2010 (UTC)[reply]

I also think that we should drop the extra. Its presence would be warranted (or even required) when kinetic energy is explained in the context of rest energy or of energy in general. I think the word is highly misplaced in the first sentence of the lead of an article about kinetic energy. Surely there must be a place in the Energy article where we can refer to KE as extra energy? DVdm (talk) 08:55, 25 June 2010 (UTC)[reply]

At Energy#Forms of energy there is a list of the different forms of energy. It is clear that kinetic energy is not the only form of energy so an object can simultaneously possess energy in two or more forms. However, I very much agree with your point that the opening sentence of the article on kinetic energy is not the place to write everything that can be written about kinetic energy. The opening sentence needs to be the simplest thing that can be accurately stated about kinetic energy. The opening sentence may be the only thing read by young readers and newcomers to science. Whether they read further will depend on how meaningful the opening sentence is. Make technical articles accessible! Dolphin (t) 12:18, 25 June 2010 (UTC)[reply]

I agree with the last statement, although I don't think that this one extra word would make the opening sentence less meaningful. As far as I'm concerned the opening sentence could go like:

• The kinetic energy of an object is the extra energy which it possesses due to its motion, in addition to other possible (non-motion related) forms of energy.

I think that "extra" begs for more, and that this specification renders moot (and effectively kills) the source requirement you brought up. For me this would be a very good opening sentence as well. DVdm (talk) 12:47, 25 June 2010 (UTC)[reply]

By the way, I think that what you just did to the lead of Potential Energy will immediately scare every young reader and newcomer to science away. DVdm (talk) 12:58, 25 June 2010 (UTC)[reply]

Note - TSRibeye, I notice that you removed the word again. Please don't do this. See my message/warning on your talk page. DVdm (talk) 17:35, 25 June 2010 (UTC)[reply]

My intent behind removing "extra" from the first sentence/initial definition is that I don't want people to be confused when they are looking for a quick definition. I was having a discussion with a friend re: kinetic energy and needed a quick reference to verify I correctly remembered my college physics. The "extra energy" jumped out at me and I thought I wasn't remembering the concept correctly. A quick web search determined "extra" was not part of the generally used definition. I didn't understand the use of "extra" until JRSpriggs explained on the talk page. Solution? I propose one of my first suggestions. Remove "extra" from the first sentence, and add a sentence or two immediately following explaining the distinction between KE vs. other (extra) energy. DVdm, my apologies. How about "The kinetic energy of an object is the energy which it possesses due to its motion. While an object may simultaneously possess multiple forms of energy Energy#Forms of energy, Kinetic energy is specific to motion." Slightly redundant, but it returns the definition to its base form while distinguishing kinetic energy from rest energy (and other forms of energy). TSRibeye (talk) 20:00, 25 June 2010 (UTC)[reply]

Sure, and the current wording is ok as well for me. But if JRSpriggs really feels better with the extra word, I propose we use the above compromize (now set in bold) and move on. DVdm (talk) 20:27, 25 June 2010 (UTC)[reply]

There is an excellent, simple definition of kinetic energy in Jain's Textbook of Engineering Physics. It says The energy possessed by a body due to its motion is called its kinetic energy. It is on page 9 and immediately precedes coverage of potential energy on page 10.

Considering we are citing Jain in the opening sentence in Potential energy I suggest we do the same in the opening sentence in Kinetic energy. Dolphin (t) 12:18, 5 July 2010 (UTC)[reply]

Apparently user User:JRSpriggs is claiming ownership of this article against WP policies by reverting obvious improvements to the lead without explanation. For examples, the article stated that the body maintains this kinetic energy unless its speed changes. This is such a naive formulation, that one must ask, why would it change its speed? This is not the only condition, and equally naive would be the statement, .. unless it changed its mass. The correct statement is that it maintains it kinetic energy until another force acts on the object (to either change mass or speed). The rest of the lede had other either confusing or specialist language in poor style. Kbrose (talk) 21:43, 16 September 2010 (UTC)[reply]

Small point: This is actually a stronger/more strictly correct formulation than the force statement. You can apply forces to accelerate an object without changing its speed. I make no comment on the targeting of the lede. User A1 (talk) 22:22, 16 September 2010 (UTC)+[reply]

Perhaps you need to look at the definition of acceleration: a = v / t. If you accelerate something, you have to change its velocity (v), or shrink time. Kbrose (talk) 22:56, 16 September 2010 (UTC)[reply]

But you dont need to change speed. Circular motion User A1 (talk) 22:58, 16 September 2010 (UTC)[reply]

I can see why JRSpriggs reverted the edits by Kbrose. Some of the changes made by Kbrose were incorrect. For example, Kbrose wrote that a body maintains its kinetic energy as long as no other force acts upon it. When a body is moving along a curved path a centripetal force is acting upon it. A centripetal force does no work on a body and so does not alter the body's kinetic energy.

The statement by Kbrose would have been correct if it said as long as no other force does work upon it. Dolphin (t) 22:53, 16 September 2010 (UTC)[reply]

Kbrose has written that acceleration a = v / t. This is true in rectilinear motion but it is not true generally. For three-dimensional motion, acceleration, velocity and displacement are all vector quantities. A particle or body moving along a curved path is undergoing an acceleration, even though its speed may be unchanged. Dolphin (t) 23:08, 16 September 2010 (UTC)[reply]

Due to the incorrect nature of these changes, I have reverted them. I am in a rush right now, so later i will come back and restore the non-wrong bits. User A1 (talk) 08:34, 17 September 2010 (UTC)[reply]

I am sorry that I did not post here earlier. I meant to explain my second reversion, but I was called away from my computer. I am not claiming ownership of this article — I have allowed many changes that I did not entirely like over the years that it has been on my watch list. But I feel that Kbrose should show more respect for the fact that many others have already polished the article.
The part of Kbrose's edit affecting the lead contained these changes which I felt were objectionable:

• He changed "unless its speed changes" to "as long as no other forces act upon it". This obscures the fact that kinetic energy is a function of speed. And implicitly raising issues about mass-change in the lead is inappropriate for the intended audience of this article who are probably at high school level or below.
• He introduced the words "the same bullet has zero kinetic energy with respect to an observer riding the bullet". This is ludicrous and will confuse or put off the reader.
• He said "at the same velocity and in the same direction". This is redundant because velocity includes direction as well as speed.
  

JRSpriggs (talk) 10:10, 17 September 2010 (UTC)[reply]

In the "Relativistic kinetic energy of rigid bodies" section, it is assumed that ${\displaystyle p=\gamma mv}$, and then the kinetic energy ${\displaystyle E_{k}}$ is derived from that, somewhat laboriously. Why not instead just start with the assumption that ${\displaystyle E=\gamma mc^{2}}$ and skip all that derivation? IMHO, the expression for ${\displaystyle p}$ is no more or less intuitive than that for ${\displaystyle E}$, so I see no advantage to starting with ${\displaystyle p}$ rather than ${\displaystyle E}$ in that sense. Also, IMHO, the derivation itself isn't that enlightening given that the more transparent, non-relativistic derivation is well discussed earlier in the article. Quantling (talk) 20:09, 24 September 2010 (UTC)[reply]

In Newtonian physics, conservation of momentum (defined as mv) is a given. Conservation of energy including kinetic energy is derived.

I think that we need to follow the same pattern in special relativity, if only to make it clear where Newtonian physics goes wrong. JRSpriggs (talk) 07:41, 27 September 2010 (UTC)[reply]

Okay, I am giving it another try with the article edits. Feel free to modify my changes. Especially if you choose to revert, please leave detailed comments here. Thanks — Quantling (talk) 18:39, 27 September 2010 (UTC)[reply]

I very much prefer the original derivation for the reasons given by JRSpriggs. This new derivation uses the expression ${\displaystyle \mathbf {v} ={\frac {\mathbf {p} c^{2}}{\sqrt {m^{2}c^{4}+p^{2}c^{2}}}}}$ which seems to be falling from the sky to only serve our purpose in the imminent derivation. Do we have a proper source for either derivation? If not, I think we should find a good source and stick with it. This is a source with a good derivation, but it uses yet another slightly different approach. Thoughts? DVdm (talk) 19:19, 27 September 2010 (UTC)[reply]

Interesting. The expressions, ${\displaystyle \mathbf {v} ={\frac {\mathbf {p} c^{2}}{\sqrt {m^{2}c^{4}+p^{2}c^{2}}}}}$ and ${\displaystyle \mathbf {p} ={\frac {m\mathbf {v} }{\sqrt {1-v^{2}/c^{2}}}}}$, are mathematically equivalent. Yet I take it that you consider the latter to be more accessible to the lay reader. I guess I could agree with that. On the flip side, the ${\displaystyle \int \mathbf {v} \cdot d\mathbf {p} }$ formulation calls out for doing the math as a function of p rather than of v, yes? (For instance, if there were yet another way to relate momentum to velocity, kinetic energy might most easily be derived by first writing v in terms of p and then integrating.) Maybe there is some way to have it both ways; perhaps by first proving the formula "that seems to be falling from the sky" and then proceeding with p. Thoughts? Quantling (talk) 20:39, 27 September 2010 (UTC)[reply]

Sure, I know they are equivalent, but I have never seen the expression in the literature. Here in Wikipedidia we need verifiable sources - see wp:V and wp:RS. We also are not allowed to create and present a quick proof here, as that would be orginal research (see wp:NOR), and note that this is not just a trivial calculation in the spirit of wp:CALC. So, unless we have a source for the expression, I'm afraid it will have to go. On the other hand, I have seen int{ v dp } been done as int{ v d(m γ v) } etc... many times in the literature, so I guess a source can easily be found for it.

Meanwhile, since throwing a {{cn}}-tag at the expression looks a bit silly, I have restored the "original" version. I suggest we keep it that way, and first discuss what we do with the section on the talk page. DVdm (talk) 21:24, 27 September 2010 (UTC)[reply]

What if we introduced the integration by parts

${\displaystyle \int \mathbf {v} \cdot d\mathbf {p} =\mathbf {p} \cdot \mathbf {v} -\int \mathbf {p} \cdot d\mathbf {v} }$

in the non-relativistic section before we plugged in ${\displaystyle \mathbf {p} =m\mathbf {v} }$? Then we could use

${\displaystyle E_{k}=\mathbf {p} \cdot \mathbf {v} -\int \mathbf {p} \cdot d\mathbf {v} }$

in the relativistic section by directly plugging in ${\displaystyle \mathbf {p} ={\frac {m\mathbf {v} }{\sqrt {1-v^{2}/c^{2}}}}\,.}$ Quantling (talk) 13:12, 28 September 2010 (UTC)[reply]

I would not want to complicate the non-relativistic section in order to simplify the relativistic section because many more people will look at the former than at the latter. JRSpriggs (talk) 14:09, 28 September 2010 (UTC)[reply]

The IPB is very nice, but whatever we do, I think it should be solidly sourced. Otherwise the section will never stabilise. DVdm (talk) 16:22, 28 September 2010 (UTC)[reply]

FWIW, ${\displaystyle H=\int \mathbf {v} \cdot d\mathbf {p} }$ is the kinetic part of a Hamiltonian, though this relationship is usually expressed as ${\displaystyle dH/dp_{i}=v_{i}}$. Similarly, ${\displaystyle L=\int \mathbf {p} \cdot d\mathbf {v} }$ is the kinetic part of a Lagrangian, though this relationship is usually expressed as ${\displaystyle dL/dv_{i}=p_{i}}$. The integration by parts to give the Hamiltonian in terms of the Lagrangian is over a century old. — Quantling (talk) 20:22, 28 September 2010 (UTC)[reply]

Afaiac, put it in the article, perhaps in a separate little section, but please make sure to include a proper source citation. Thx - DVdm (talk) 20:55, 28 September 2010 (UTC)[reply]

Assume, that for moving some mass m metal box on wheels or without wheels on some surface (it can be wood or metal or plastic) 1 m distance is used the same amount of energy as accelerating same mass m metal box in cosmos (where no friction) to speed 1 m/s. Assume that in both cases there no air friction. For example, there is metal box of mass m=1 kg. To move this metal box on horizontal ice surface need the same amount of energy as accelerate this metal box in cosmic space without gravity [by, say, hand of astronaut] to the speed of 1 m/s. So any object accelerated in cosmic space to speed v have energy W=m*v (please, don't fak my brains with Einshtein relativity and over nonsenses). So then total kinetic energy of moving metal box on ice surface and accelerated to speed v is ${\displaystyle W_{total}=m\cdot v+m\cdot S=m\cdot v+m\cdot t\cdot {\frac {v}{2}}}$ . Where v/2 is average speed of metal box (and v is final speed of metal box). Metal box on ice surface was accelerated to speed v=1000 m/s in 100 seconds. So then acceleration is a=v/t=1000/100=10 (m/s)/s. So total kinetic energy for accelerating metal box on ice surface is ${\displaystyle W_{total}=m\cdot v+m\cdot S=m\cdot v+m\cdot t\cdot {\frac {v}{2}}=1\cdot 1000+1\cdot 100\cdot {\frac {1000}{2}}=1000+50000=51000}$ (some energy units). And to accelerate metal box of mass 1 kg in cosmic space without gravity need energy W=m*v=1*1000=1000 (some energy units). So do you see difference how much need kinetic energy in cosmic space and on earth because of friction into surface of earth, huh?

Example number 2:

Metal box mass is m=1 kg. Metal box is accelerated with acceleration 10 (m/s)/s to the speed 100 m/s. Metal box accelerated to 100 m/s after 10 s. This all is correct, because v=a*t=10*10=100 m/s. So need to find kinetic energy of metal box on ice accelerated to speed 100 m/s and metal box kinetic energy accelerated in cosmic space without gravity. In both cases air friction is ignored.

Kinetic energy in cosmic space is W=m*v=1*100=100. And kinetic energy on ice surface is W=m*v+m*t*v/2=1*100+1*10*100/2=100+500=600.

Remember, this is because need the same amount of energy to push 1 kg mass object on surface with gravity at distance 1 meter like need the same amount of energy to accelerate 1 kg mass object to speed 1 m/s in cosmic space.

All examples was calculated without air friction. But I have to say something about air friction. Nobody really knows how air friction working... It can be, that if assume in cosmic space is air friction like on Earth, then there is possible two variants of calculating air friction for object of mass m. Either the same way as calculating friction into surface, then to overcome this friction need energy W=k*t*v/2, where k is some coefficient depending on mass and object surface area, but for 1 kg metal cube k=~0.001. Or another way to calculate air friction is W=k*4*(S*v/2)/3=k*4*S*v/6, where S is distance, which was moved 1 kg metal cube, a is acceleration, v is speed to which was accelerate 1 kg metal cube and k is coefficient k=~0.001.

I can explain why in square increasing energy for air friction, when speed increasing need formula W=k*4*(S*v/2)/3=k*4*S*v/6 and not W=k*(S*v/2)=k*S*v/2. It is because of calculation through many sums... For example a=10 m/s/s, t=40 s. Then v=a*t=10*40=400 m/s; S=v*t/2=a*t*t/2=400*40/2=16000/2=8000 m. Energy waisted for air friction W=S*v/2=8000*400/2=1600000. But according calculation with many small parts:

W=5*5+15*15+25*25 +35*35*45*45+55*55+65*65+75*75 +85*85+95*95 +105^2+115^2+ 125^2+135^2 + 145^2+ 155^2+ 165^2+ 175^2+185^2 +195^2+205^2 +215^2+225^2 +235^2 +245^2+255^2 +265^2 +275^2+285^2+295 + 305^2+315^2 +325^2+335^2 +345^2+ 355^2+365^2+375^2+385^2+395^2 = 25+225+625+1225+ 2025+3025+4225+5625 +7225+9025 +11025+...+156025= 33250 +2099750 =2133000. So 2133000/1600000=1.333125=4/3, so this is explanation from where comes 4/3. If you wondering what are those numbers 5*5, 15*15, 25*25 and so on, then I can explain it, ...you see, if acceleration is a=10 m/s/s, then per first second average moving distance is (0+10)/2=5 m, per second second average moving distance is (10+20)/2=15 m, but number 15 is also average speed 15 m/s. So it's just like Changing formula S*v/2 to ${\displaystyle s_{1}\cdot v_{1,average}+s_{2}\cdot v_{2,average}+s_{3}\cdot v_{3,average}+s_{4}\cdot v_{4,average}+...+s_{40}\cdot v_{40,average}.}$

Also need to be very cautious by choosing coefficient k for second air friction calculating formula (k*4*S*v/6, when air friction increasing square), because can be some things which I didn't sow yet and coefficient can be then more like k=0.000001.

And if you wondering how to calculate air friction, when object moving with constant velocity, then method in first (not square air friction depnding on speed) is simple W=k*S=k*t*v/2, where S is distance which object was moved; t is time, which object was traveling, v is speed at which object was traveling and k=~0.001.

And if air friction is much bigger at bigger speed (with formula k*4*S*v/6), then need calculate this way: W=k*S*v, and then need very carefully choose coefficient k and v=constant.

And friction on ice surface or in bike wheels always calculated with formula W=k*m*S=k*m*t*v at constant speed v=constant. Only if for ice surface friction coefficient k=1, then for bike wheels friction k=~0.5.

—Preceding unsigned comment added by 84.240.9.58 (talk • contribs) 23 October 2010

"So any object accelerated in cosmic space to speed v have energy W=m*v " does not make sense.--Patrick (talk) 10:28, 23 October 2010 (UTC)[reply]

I will tell you what don't make sense, in cosmos no friction, so why kinetic energy is not W=F*t=m*a*t ? It's about how long you acting the same force on object and not about what distance you force acting on object: W=m*a*S.

"As a result of this, a car traveling twice as fast requires four times as much distance to stop (assuming a constant braking force. See mechanical work)." Do you really believe, that car moving at 40 m/s will take braking time 4 times longer than car moving at speed 20 m/s ? It's only accelerating will take about 4 times more energy accelerate to 40 m/s, than to 20 m/s, because of friction, thats why experiments give such results. Because will be traveled 4 times longer distance. Look, a=3 (m/s)/s, ${\displaystyle v_{1}=40}$ m/s, ${\displaystyle v_{2}=20}$ m/s, v=a*t, ${\displaystyle t_{1}={\frac {v_{1}}{a}}={\frac {40}{3}}=13.33333}$ s; ${\displaystyle t_{2}={\frac {v_{2}}{a}}={\frac {20}{3}}=6.66667}$ s. Distance S=t*v/2; v is final speed. ${\displaystyle S_{1}=t_{1}\cdot {\frac {v}{2}}={\frac {13.33333\cdot 40}{2}}=266.66667}$ m; ${\displaystyle S_{2}=t_{2}\cdot {\frac {v}{2}}={\frac {6.66667\cdot 20}{2}}=66.66667}$ m. And 266.6666/66.6666=4. That's why. It's friction, thats why experiments are such. It's not of kinetic energy, it's of friction. In real life braking will take only 2 times longer time braking.