I Kinetic Energy derivation assumption?

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The discussion centers on the assumption that the classical kinetic energy equation, KE=(1/2)m*v^2, is derived under the condition of constant acceleration. Participants question why many textbooks and authors emphasize this assumption, despite the ability to derive the kinetic energy equation using time-varying acceleration. It is suggested that this simplification is often made for educational purposes, as it provides a clearer entry point into the concept of kinetic energy. Additionally, the conversation touches on the validity of the kinetic energy equation in more complex scenarios, such as relativistic motion, where the equation still holds under specific conditions. The consensus indicates that while the constant acceleration assumption is useful for beginners, it does not limit the broader applicability of kinetic energy principles.
Roboto
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The classical definition to the Kinetic Energy equation is KE=integral of F*dx where F=d(m*v)/dt. When mass is constant, KE=(1/2)m*v^2.

I am working on a vibration problem at work and having to review my Lagrangian Dynamics books from 30 years ago. So my question is about all of the authors of these books (and internet searches) make the same comment that this assumes that acceleration is constant. They make no further statement other than for simplicity they will use the classical definition for the Kinetic energy equation.

Why would they make that comment? When i use accelerations that are a function of time (and mass being constant) i can derive the same general equation form where kinetic energy is now just a function of time.

Though this is not a stumbling point for me, i am curious why all the authors would make that side comment that kinetic energy equation assumes that the acceleration being a constant.
 
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Roboto said:
So my question is about all of the authors of these books (and internet searches) make the same comment that this assumes that acceleration is constant. They make no further statement other than for simplicity they will use the classical definition for the Kinetic energy equation.
Why would they make that comment? When i use accelerations that are a function of time (and mass being constant) i can derive the same general equation form where kinetic energy is now just a function of time.
Though this is not a stumbling point for me, i am curious why all the authors would make that side comment that kinetic energy equation assumes that the acceleration being a constant.
Because it's easier for the entry-point into the concept?
 
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I don't understand the statement that the derivation of the "work-energy theorem" were only valid for constant acceleration. Where have you read this?

It's very easy to prove: Just define the kinetic energy
$$T=\frac{m}{2} \dot{\vec{x}}^2$$
and take its time derivative and use the equation of motion,
$$\dot{T}=m \dot{\vec{x}} \cdot \ddot{\vec{x}}=\dot{\vec{x}} \cdot \vec{F}.$$
If now ##\vec{F}## has a time-indepnendent potential,
$$\vec{F}=-\vec{\nabla} V(\vec{x})$$
you get
$$\dot{\vec{x}} \cdot \vec{F} = - \dot{\vec{x}} \cdot \vec{\nabla} V = -\mathrm{d}_t V(\vec{x}),$$
and then the work-energy theorem simplifies to
$$\dot{T}=-\dot{V} \; \Rightarrow \; \dot{E}=\mathrm{d}_t (T+V)=0 \; \Rightarrow \; E=\text{const},$$
i.e., energy conservation.
 
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vanhees71 said:
i.e., energy conservation.
In a conservative system, the Energy supplied must equal the Kinetic Energy out. However the Work is done, you have to get the same answer or we have a 'loophole' and the makings of a PPM.
 
Yeah, I understand the conservation of energy, and the derivations. Easy peasy.

What I haven't figured out is why would my textbook authors, along with searches on the internet, have this comment that the derivation of the kinetic energy equation is based on the assumption that acceleration is constant.

So i thought i would bounce this off the group if they knew of cases where derivation kinetic energy would not be (1/2)m*v^2 if acceleration was not constant.
 
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I've never read this nonsensical statement anywhere. There are of course some overly "didactical" high-school textbooks out there, which confuse the students more than helping them learning physics ;-).
 
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Roboto said:
What I haven't figured out is why would my textbook authors, along with searches on the internet, have this comment that the derivation of the kinetic energy equation is based on the assumption that acceleration is constant.
When I googled "deriving kinetic energy equation", the top hit was:

https://physics.info/energy-kinetic/

Which does exactly as you say. As did the first video hit. Although the second video hit used more general acceleration.

Interesting. I suggest these are elementary introductions for those who haven't yet encountered non-constant acceleration.
Roboto said:
So i thought i would bounce this off the group if they knew of cases where derivation kinetic energy would not be (1/2)m*v^2 if acceleration was not constant.
Well the relativistic kinetic energy is ##(\gamma -1)mc^2##, which is approxaimately ##\frac 1 2 mv^2## when ##v << c##.
 
vanhees71 said:
I've never read this nonsensical statement anywhere. There are of course some overly "didactical" high-school textbooks out there, which confuse the students more than helping them learning physics ;-).
Yes: “confuse” is the word. Whether or not Mr Worthington actually said “constant acceleration”, he didn’t stress it. All the SUVAT equations he derived used straight lines (of course) but, despite him being a lovely old guy, he disturbed my sleep with that cognitive dissonance. All he need have said was “ assuming ……..”
 
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PeroK said:
When I googled "deriving kinetic energy equation", the top hit was:

https://physics.info/energy-kinetic/

Which does exactly as you say. As did the first video hit. Although the second video hit used more general acceleration.

Interesting. I suggest these are elementary introductions for those who haven't yet encountered non-constant acceleration.

Well the relativistic kinetic energy is ##(\gamma -1)mc^2##, which is approxaimately ##\frac 1 2 mv^2## when ##v << c##.
I don't think that it is a good strategy to use some movies or other arbitrary material from a Google search. Reliable are usually manuscripts accompanying lectures at universities and, of course, proper textbooks. If you insist on "movies" it's much less easy to give a general advice. You find some very good ones on youtube and in these "Covid times" also very many "online lectures", which also are usually good, but it is not surprising that on youtube there are also many bad ones. So, if you learn something new, so that you cannot easily judge yourself whether a source is reliable or not, it is important to cross-check with more reliable sources, i.e., lecture notes or, usually even better, textbooks and of course original scientific publications in peer-reviewed reliable journals. If the subject is not too new, you also find review articles as in journals like Review of Modern Physics or Physics Reports. For classical mechanics, of course, there are many good textbooks (the only exception is the >=3rd edition of Goldstein, which was spoiled by co-authors, while the 1st and 2nd editions are standard references).
 
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My Lagrangian Dynamics textbooks from 30+ years ago had the comment that the derivation of the classical Kinetic Energy equation assumes acceleration is a constant. I just don't understand why multiple authors would make comments that the derivation assumes acceleration is a constant.
 
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The derivation of the classical kinetic energy equation does not assume anything like that, as show in #3. You can as well derive why the Lagrangian of a system of interacting particles looks the way it looks, using the full Galilei symmetry of Newtonian spacetime and Noether's 1st theorem.
 
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Roboto said:
I just don't understand why multiple authors would make comments that the derivation assumes acceleration is a constant.
That 'elementary' assumption is ok if you bolt on the requirement that energy is conserved. Then any profile of acceleration will give the same result. It's fair enough for a first time round the topic - same as a lot of entry level Physics.
 
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