Answer: E = mv2: Understanding Anti Newton's Equation

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In summary: Energy (E) = WorkE = FsE = masE = msv/t --------(because a =v/t)E = mvs/tBut s/t = v.: E = mv2 -----------(1)But according to Isaac Newton E = mv2/2There are several problems here, all of which stem from ignoring calculus.
  • #1
Lord Advait
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Energy (E) = Work
E = Fs
E = mas
E = msv/t --------(because a =v/t)
E = mvs/t
But s/t = v
.: E = mv2 -----------(1)
But according to Isaac Newton E = mv2/2
 
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  • #2
Lord Advait said:
Energy (E) = Work
E = Fs
E = mas
E = msv/t --------(because a =v/t)
E = mvs/t
But s/t = v
.: E = mv2 -----------(1)
But according to Isaac Newton E = mv2/2

This is what happens when you simply plug in things without understanding what they mean.

Notice that when you use "a=v/t", you have already made an assumption that this is a VARYING "v", and that this is really some average value of a uniformly varying v. Yet, you used "s/t=v", which would only work for a CONSTANT v. Just think about it, if v is changing and getting larger when there's an acceleration, what value did you just compute using s/t = v? Which "v" is this when it is a changing value?

Zz.
 
  • #3
But s/t = v

What happened to the acceleration?

Edit: Late as usual. Opened this thread and went to browse other places in the web. :D
 
  • #4
Why does everyone want to pick on good old Newton these days? :smile:
 
  • #5
Lord Advait said:
Energy (E) = Work
E = Fs
E = mas
E = msv/t --------(because a =v/t)
E = mvs/t
But s/t = v
.: E = mv2 -----------(1)
But according to Isaac Newton E = mv2/2

There are several problems here, all of which stem from ignoring calculus.

The work energy theorem, when stated correctly says:

[tex]E = \int \vec{F} \cdot d\vec{s}[/tex]

This only takes the form you stated when [tex]\vec{F}[/tex] is constant and the path traveled is in the same direction [tex]\vec{F}[/tex] points.

Replacing [tex]\vec{F}[/tex] by [tex]m\vec{a}[/tex] is only valid if what you're considering is the net force on the object, not one force among many. If it is the net force, this step is fine.

However, the next step is not. Acceleration is the rate of change of velocity:

[tex]\vec{a} = \frac{d\vec{v}}{dt}[/tex]

Even if you only care about the average acceleration, you need the change in velocity. The only situation where you can write [tex]\vec{a} = \frac{\vec{v}}{t}[/tex] is when [tex]\vec{v} = 0[/tex] at time 0.

The same sort of considerations apply when talking about velocity:

[tex]\vec{v} = \frac{d\vec{s}}{dt}[/tex]

But, now we have the worse problem that you've already assumed that [tex]\vec{v}[/tex] isn't constant; so there's no situation where you can just write [tex]\vec{v} = \frac{\vec{s}}{t}[/tex].

Finally, Newton never said anything about energy, much less that [tex]E = \frac{1}{2} mv^2[/tex]. He believed that momentum told you everything you needed to know. It was Leibniz who first introduced the idea of energy (although he called is "vis viva").

Here is the quickest way to get the form for kinetic energy out of the work energy theorem algebraically. I will need to assume that the force discussed is the net force on the object, that the net force is constant, and that the force acts along the direction of motion.

[tex]\Delta E = F\Delta x[/tex]

[tex]\Delta E = ma\Delta x[/tex]

[tex]\Delta E = m \frac{\Delta v}{\Delta t} \Delta x[/tex]

[tex]\Delta E = m \Delta v \frac{\Delta x}{\Delta t}[/tex]

[tex]\Delta E = m (v_f - v_i) \overline{v}[/tex]

[tex]\Delta E = m (v_f - v_i) \frac{v_f + v_i}{2}[/tex]

[tex]\Delta E = \frac{1}{2} m (v_f^2 - v_i^2)[/tex]

[tex]\Delta E = \frac{1}{2} m \Delta v^2[/tex]

[tex]\Delta E = \Delta \left (\frac{1}{2} mv^2\right )[/tex]
 
  • #6
excellent parlyne...good proof
 
  • #7
radou said:
Why does everyone want to pick on good old Newton these days? :smile:

I don't know, but it is getting quite annoying. .
 
  • #8
radou said:
Why does everyone want to pick on good old Newton these days? :smile:
It's a homophobic reaction, I guess.
 

Related to Answer: E = mv2: Understanding Anti Newton's Equation

1. What is E = mv2?

E = mv2 is a mathematical equation that is often referred to as the "anti Newton's equation". It is a variation of Newton's second law of motion (F = ma) and is used to calculate the kinetic energy of an object in motion.

2. How is E = mv2 different from Newton's second law?

While Newton's second law (F = ma) relates an object's mass and acceleration to the net force acting on it, E = mv2 relates an object's mass and velocity to its kinetic energy. This equation is often used in situations where the force acting on an object is not constant.

3. What does the "anti" in "anti Newton's equation" mean?

The term "anti" in this context refers to the fact that E = mv2 is essentially the opposite of Newton's second law. While Newton's second law focuses on the relationship between force and acceleration, E = mv2 focuses on the relationship between mass and velocity.

4. How is E = mv2 used in real life?

E = mv2 is used in a variety of fields, including physics, engineering, and sports. It is commonly used to calculate the kinetic energy of moving objects, such as projectiles or vehicles. It is also used in the design of roller coasters and other amusement park rides.

5. Is E = mv2 always accurate?

Like any mathematical equation, E = mv2 is only accurate when used in the right context and with the correct input values. It is based on certain assumptions and may not be applicable in all situations. It is important to understand the limitations and applicability of this equation before using it in any calculations.

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