Why do you think work is given by 2Fd? Do you want it to be that? Do you want to define work as two times the force times the distance the force acted over?
You can, if you like, but you'll be plagued by a factor of 2 in every place where you will talk about any work.
It's way easier to keep this factor of 1/2 just in definition of kinetic energy instead.
You can name mv^2 kinetic energy, and 2Fd work. But then you can name (mv^2)/2 kinetic half-energy and Fd half-work and physicist will still prefer to do calculations for half-work and half-energy because there will be less places where this factor of 2 will crop up in calculations.
Probably about many things in his life as we all are.
> The work done has nothing to do with distance. It only has to do with velocity.
The thing that has to do only with the velocity we call kinetic energy.
Work is something else.
> Accelerating the body from 0 to a final velocity is the work.
It's one example of work. You can also consider for example a force acting upon a body that travels with constant velocity of v over a distance d. And this force would still do the work equal to Fd. You might ask, how is it possible that this body that the force is acting upon doesn't have an acceleration? Because at the same time this body might have interactions with other bodies through different forces and do work on them.
As example of this imagine a body that moves directly up on Earth with constant velocity v, because the force that does the work on it is directed up and equal exactly mg. That force does the work on this body, but none of it has anything to do with velocity and it goes directly into potential energy.
Regardless of what kind of motion the body uses to travel distance d, if it's under the influence of the force F during that, the equation holds that work done by the force F equals Fd.
This allows us to do calculations for various kinds of energies and multiple acting forces, just adding the work that each force does and changes in every kind of energy the body has.
> If you know the distance and the acceleration, you can use them to calculate the final velocity with the well-known kinematics equation v² = 2ad.
This equation is not some core law of the universe. It just comes from assuming we have a movement under influence of a single force of a body that at t=0 started at rest. It comes from v=at and d=(at^2)/2 and those come from integrals over time of a done once and twice.
This equation you really like is just a result of how we define speed and acceleration simply applied to one kind of motion. It has nothing to do with forces or energies. It just deals with displacement and its first and second derivatives (velocity and acceleration).
> Likewise, if you know the elapsed time and the acceleration you can use the formula v = at to get the final velocity. It's simple, really.
Yes. It's simple to the point of not being especially interesting.
What's interesting is that resulting formula deltaEk = Fd is more general. It works regardless of whether the motion that causes the change in kinetic energy has constant acceleration or not. And you can use it when multiple forces act upon this body and multiple kinds of energies are involved, for example deltaEp + deltaEk = Fd + Ge + Hi (where deltaEp is a change in potential energy, deltaEk is a change in kinetic energy, F, G, H are the forces and d, e, i are the displacement of the body while the forces F, G and H acted upon the body respectively).
You are not wrong writing mv^2 = 2Fd and we might have called mv^2 energy but the we would have to call 2Fd work, and we really prefer to call Fd work because that factor 2 would crop up everywhere in physics where there are any energies and works considered (you already saw it in your version of the potential energy equation Ep=2mgh, and there are so many places in physics where it would need to show up, basically any place where there are energies and fields involved). So it's just more convenient to have it in the denominator of kinetic energy than literally everywhere else.
> It's simple to the point of not being especially interesting.
Either you don't realize what you're saying or you are playing a game of deception.
1. This equation is indeed a core law of the universe.
2. It is extremely interesting because it has a specific meaning. It expresses the kinetic energy of a massive body in motion.
3. More specifically, it means that E = mc² does not represent what Einstein claimed it did. It represents the maximum kinetic energy that a massive body can have. This is why it is extremely interesting.
Thank you for the exchange.
I'm sorry you feel that way. I'm just trying to help you understand source of those equations.
> 1. This equation is indeed a core law of the universe.
Let me show you step by step where the equation v² = 2ad comes form.
Let's consider an object moving in any way. It may have some mass, it may be massless, it doesn't matter.
Let's call the distance it moved so far d. Because object moves, d is a smooth function of time so we might be interested with its first and second derivative. Let's call the first derivative v and second derivative a. Those might be constant or might be also some functions of time.
Then let's say we are interested in a very specific type of motion. A motion where at t=0 first derivative was equal zero, and second derivative is constant through the whole motion.
So we have second derivative equal some constant value a. Because integration is the opposite of derivation to calculate v we need to integrate a over time t. This is a purely mathematical operation. And gives us v = at Then we want to calculate the function d itself for this specific motion. To do that we need to integrate over time again and we get d = at²/2 This again is purely mathematical operation that doesn't rely on any connection to the world we live in.
So we have:
d = at²/2
v = at
when we calculate t from the second equation and put it in the first one we get:
d = av²/2a²
by rearranging we get:
v² = 2ad
All of the above is true and pure math. Even if universe didn't exist or worked completely differently (for example if real objects didn't really move in smooth motions but teleported instead) this equation and its derivation would still exist, it just would not reflect reality.
So this equation comes from pure math and a concept of how movement might work, defined by purely mathematical means. It doesn't have to describe our universe and existence of this equation doesn't depend on the existence of our universe or how movement actually works in our universe. So it's not a core law of universe, just a mathematical consequence of considering one specific type of motion that we think might happen in our universe.
> It is extremely interesting because it has a specific meaning. It expresses the kinetic energy of a massive body in motion.
Yes, but that's all. There are many other forms of energy. And what's way more interesting, that we get from considering kinetic energy, is the concept of energy itself, that it can come in many forms and can get transferred from one form to another and that transfer is mediated by work done by forces acting on moving bodies over some distances.
> 3. More specifically, it means that E = mc² does not represent what Einstein claimed it did. It represents the maximum kinetic energy that a massive body can have. This is why it is extremely interesting.
Absolutely not. Kinetic energy defined as mV² (with 1/2 or without) is only the approximation what the kinetic energy of a massive body is when it travels at slow speeds. When the speeds get near c best approximation of kinetic energy that we know is
Ek = (1/sqrt(1-v²/c²) - 1)mc²
It's not immediately obvious but this equation can be approximated by mv²/2 for small v (by using Maclaurin series expansion of part of it and taking only first terms).
This better equation again is purely mathematical result derived by considering of how adding movements must work if there is such a thing as the maximum speed of movement (that we called c).
You can see that this equation tends to infinity as v approaches c. So there's no maximum kinetic energy. You can pump energy into the moving body by applying the force to it to do the work regardless of what speed the body has already. Buy pumping in more energy you just bring the speed closer to c but you never reach it no matter how much you pump in.
Maximum kinetic energy doesn't make sense conceptually. If there was such thing then by putting in the work to accelerate the body, at some point, all the work you put in so far would reach this maximum kinetic energy. This would need to happen at speed below c because there's no amount of force you can apply (over whatever time) to make massive object travel at exactly c.
What then? When you already reached maximum kinetic energy, what if you would still try to put in more work? Where would the work go then if not into kinetic energy?
> Thank you for the exchange.
I have some fringe opinions about some aspects of physics as well and I would very much like if someone with more intimate knowledge of math involved would point out exactly what's wrong with them.