4 ms·

> Not every value is divided by 2. Only the K.E. term is divided by 2. The P.E. term is not. I'm not sure what that means. > If you ever find where Young work

> Not every value is divided by 2. Only the K.E. term is divided by 2. The P.E. term is not. I'm not sure what that means. > If you ever find where Young worked out the equivalency to P.E. then you'll find he had m v² = 2 m g h. I don't get it. mv² is not equal to 2mgh. That makes no sense.

SirIsaac 5y ago
> mv² = 2mgh I think I now know what you mean. You mean that a mass will gain kinetic equal to 1/2 mv² by the time it hits the ground from a height of h. This doesn't prove your claim. Using my own derivation of kinetic energy, I get mgh = mv². Here's why: PE = mgh is the same thing as Newton's Work equation, W = Fd = mad. Why? Because g is acceleration (a) and h is distance (d).
- eesmith 5y ago
  mgh = mv² is experimentally disproven in the ballistics lab experiments I mentioned.
  - SirIsaac 5y ago
    I don't believe it. They're probably measuring mass and velocity and applying the formula afterwards. That doesn't prove anything. Back in Leibniz' days, they would use soft clay and measure the depths of the indentations made by falling weights. This was accurate enough to tell them that the kinetic energy was proportional to mv².
    - eesmith 5y ago
      Your "probably" means you've refused to look at the relevant experimental evidence? "proportional to mv²" also means "proportional to 1/2 mv²" and "proportional to 123.45 mv²". Assuming your gravitational and inertial masses are supposed to be the same, then 1/2 m v² is, by Noether's theorem, a consequence of conservation of energy. https://en.wikipedia.org/wiki/Noether%27s_theorem#Example_1:_Conservation_of_energy https://en.wikipedia.org/wiki/Noether%27s_theorem#Example_1:... .
      - SirIsaac 5y ago
        >"proportional to mv²" also means "proportional to 1/2 mv²" and "proportional to 123.45 mv²". Yes. This is my point. The error has never been found experimentally because everything eventually gets resolved to Newton's E = Fd. Dividing every result by 2 makes no difference because the proportions do not change. I still don't see why gravitational mass is a problem but I'll keep looking into it.

SirIsaac 5y ago

> mv² = 2mgh I think I now know what you mean. You mean that a mass will gain kinetic equal to 1/2 mv² by the time it hits the ground from a height of h. This doesn't prove your claim. Using my own derivation of kinetic energy, I get mgh = mv². Here's why: PE = mgh is the same thing as Newton's Work equation, W = Fd = mad. Why? Because g is acceleration (a) and h is distance (d).

eesmith 5y ago
mgh = mv² is experimentally disproven in the ballistics lab experiments I mentioned.
- SirIsaac 5y ago
  I don't believe it. They're probably measuring mass and velocity and applying the formula afterwards. That doesn't prove anything. Back in Leibniz' days, they would use soft clay and measure the depths of the indentations made by falling weights. This was accurate enough to tell them that the kinetic energy was proportional to mv².
  - eesmith 5y ago
    Your "probably" means you've refused to look at the relevant experimental evidence? "proportional to mv²" also means "proportional to 1/2 mv²" and "proportional to 123.45 mv²". Assuming your gravitational and inertial masses are supposed to be the same, then 1/2 m v² is, by Noether's theorem, a consequence of conservation of energy. https://en.wikipedia.org/wiki/Noether%27s_theorem#Example_1:_Conservation_of_energy https://en.wikipedia.org/wiki/Noether%27s_theorem#Example_1:... .
    - SirIsaac 5y ago
      >"proportional to mv²" also means "proportional to 1/2 mv²" and "proportional to 123.45 mv²". Yes. This is my point. The error has never been found experimentally because everything eventually gets resolved to Newton's E = Fd. Dividing every result by 2 makes no difference because the proportions do not change. I still don't see why gravitational mass is a problem but I'll keep looking into it.