Re: [free_energy] Re: E!=mgh for the Earth

Expand Messages
• ... *Which* proof, for Chrissakes? I ve lost count! Here. I ll do it again just for the exercise. We ll take teeny baby steps. Pay close attention. We will
Message 1 of 106 , Oct 1, 2004
David Thomson wrote:
> Hi Phil,
>
>
>>Several people on this list have proved that E=.5mv^2 several different
>>ways. You just refuse to accept that they're right and you're wrong.
>
>
> Point to the proof, Phil.

*Which* proof, for Chrissakes? I've lost count!

Here. I'll do it again just for the exercise. We'll take teeny baby
steps. Pay close attention.

1. F = m*a [force is mass times acceleration - Newton's second law of
motion]

2. v = a*t [velocity is acceleration times time]

3. d = v*t [distance traveled is velocity times time, for constant velocity]

4. E = F*d [energy is force times distance]

5. Energy is conserved [first law of thermodynamics]. So the kinetic
energy added to an object by accelerating it with a force is equal to
the energy expended on the object by the accelerating force.

I will first derive a formula for the total distance d traveled by an
object that starts at rest at time 0, and is then placed under constant
acceleration a. We want to know the distance traveled at time t. Since
the velocity continually increases, we can't just use equation 3 as is.
We have to take equation 2, the velocity as a function of time, and
integrate it over time.

d = definite integral from 0 to t [v(T)] dT
= definite integral from 0 to t [a * T] dT
= a * definite integral from 0 to t [T] dT [ok because a is constant]
= a * 1/2 * t^2
= 1/2 * a * t^2

[I know integral calculus is far beyond your limited mental abilities,
Dave, but it *is* the correct way to solve this problem. Besides, this
particular integral is listed in all the math tables, *and* this
specific formula is listed in the physics books. If you don't know
calculus, learn it. If you *still* don't believe this answer, repeat
Galileo's experiment and time a bunch of object falling different
distances to the ground.]

Now let's go back to the specific problem. Starting with equation 4, the
energy expended on accelerating the object, which becomes the object's
kinetic energy because it starts with none:

E = F*d

substituting equation 1 for F:

E = m*a*d

substituting our derived equation for d, and rearranging:

E = m * a * 1/2 * a * t^2
= 1/2 * m * a^2 * t^2
= 1/2 * m * (a*t)^2

But according to equation 2, a*t is just the final velocity, v, because
a is constant. Substituting v for at:

= 1/2 * m * v^2

QED.

I guess your problem is really with integral calculus, because that's
where that pesky factor of 1/2 first appears. Tough. Learn calculus.
Accept it. Ignorance of the law is no excuse.

--Phil
• ... Good point; the kinetic energy of an object *does* depend on the inertial reference frame you use to measure the object s velocity. That s true in both
Message 106 of 106 , Oct 4, 2004
mintowheel wrote:

> Part of Thomson's confusion about kinetic energy perhaps relates to
> the need to define a reference frame when assigning the specific
> value of an object's kinetic energy. Of course we must use the same
> reference frame for both the velocity and kinetic energy (an obvious
> fact to most of us). When I tried to explain this to him early on,
> he seemed unable to comprehend this, and immediately rejected my

Good point; the kinetic energy of an object *does* depend on the
inertial reference frame you use to measure the object's velocity.
That's true in both Newtonian and Einsteinean relativity. For objects on
or near the earth, the obvious choice is the earth-centered-earth-fixed
reference frame. That's the one in which the earth is stationary, though
strictly speaking it's not inertial because of the earth's rotation.

Away from the earth, things can get confusing. So you just state that
whenever you measure kinetic energy, you have to specify the reference
frame you're using.

Perhaps Dave, in his confusion, felt that kinetic energy should only be
defined in the context of a collision because the energy released in
such a collision would be invariant to an observer's reference frame.
But that's giving him way too much credit.

Phil
Your message has been successfully submitted and would be delivered to recipients shortly.