3 ms·

Coriolis was wrong. The work done has nothing to do with distance. It only has to do with velocity. Accelerating the body from 0 to a final velocity is the work

Coriolis was wrong. The work done has nothing to do with distance. It only has to do with velocity. Accelerating the body from 0 to a final velocity is the work. This is what force does. 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. That is all. 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.

scotty79 5y ago
> Coriolis was wrong. 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.
- SirIsaac 5y ago
  > This equation is not some core law of the universe. > 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.
  - SirIsaac 5y ago
    PS. Downvoting my comment is a sign of fear and cowardice. Edit: I forgot to laugh. hahahaha...HAHAHAHA...hahahaha
    - scotty79 5y ago
      I'm not sure who downvoted you. I can't downvote comments that are responses to my comments. I don't believe person who did this was motivated by cowardice. More likely annoyance or exhaustion. Or just wanted to express that he thinks you are wrong but wouldn't bother with writing a comment. Let's stick to math and physics though.
      - SirIsaac 5y ago
        I'm done here. Thanks for your input.
        scotty79 5y ago
        Please try to understand what I wrote in the previous long post. Once you do you'll be able to move on from your infatuation with mv². To understand it you'll need to have ability to calculate derivatives and integrals.
  - scotty79 5y ago

scotty79 5y ago

> Coriolis was wrong. 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.

SirIsaac 5y ago
> This equation is not some core law of the universe. > 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.
- SirIsaac 5y ago
  PS. Downvoting my comment is a sign of fear and cowardice. Edit: I forgot to laugh. hahahaha...HAHAHAHA...hahahaha
  - scotty79 5y ago
    I'm not sure who downvoted you. I can't downvote comments that are responses to my comments. I don't believe person who did this was motivated by cowardice. More likely annoyance or exhaustion. Or just wanted to express that he thinks you are wrong but wouldn't bother with writing a comment. Let's stick to math and physics though.
    - SirIsaac 5y ago
      I'm done here. Thanks for your input.
      - scotty79 5y ago
        Please try to understand what I wrote in the previous long post. Once you do you'll be able to move on from your infatuation with mv². To understand it you'll need to have ability to calculate derivatives and integrals.
- scotty79 5y ago