Work and Energy
Joel Merritt
Work is closely associated with energy, because work can be defined as a
change in kinetic energy (KE) or movement of a mass. Sir Isaac Newton told
us that work equals FD where D is the distance an object is moved directly
against a force of F. Newton also believed that the amount of motion (KE) in
a moving mass is MV where the mass is moving at velocity V. Newton cannot
have been correct in both of these equations, because these two equations
(W=Fd and KE=mv) contradict one another. Newton's work equation was accept
as correct while the energy equation was rejected. I believe that Newton's
work equation is incorrect and the KE equation was correct. Now let me
explain why I believe this.
Both the work equation and the KE equation bothered me, because some
things didn't add up about them.
1. Suppose a space ship is accelerating through space at a rate of 10
ft/sec/sec. The ship would be using nearly a constant amount of fuel, but
would be doing much more and much more work (supposedly). The work being
done by the ship would be increasing directly with the velocity of the ship
which varies directly with time.
2. Now suppose the same space ship is under the influence of gravity.
The ship is burning the same fuel as before, or using the same amount of
"energy", but if the force of gravity is equal to 10 ft/sec/sec then the
ship would be stationary and (supposedly) doing no work.
3. Suppose a bowling ball is dropped from a skyscraper. The work that
gravity would be doing to the object in accelerating it would vary directly
with time (supposedly). That means that gravity does more and more work for
every unit of time as an object falls .
4. The power of gravity would also be increasing.
All four of these past problems can be solved by defining work as force
times time (W=Ft) rather than force times distance. If this different
equation is not correct then it is a very coincidental thing for it to solve
all of these difficulties. First of all, there shouldn't be many, if any
difficulties with the correct equation. Second of all, there should be many
difficulties with an incorrect equation (such as W=Ft if it is incorrect).
When I came up with the W=Ft equation, I had no idea it would solve the
bunch of problems with the W=Fd equation. It was only because of one of the
problems that I thought W=Ft made more sense (it was quite a coincidence [?]
when it solved my other logic problems).
Kinetic Energy (KE) is today defined as .5mv^2 (one half mass times
velocity squared), but is that correct? I will give a few reasons why I
question it.
1. Suppose an object accelerates (in a Galilean universe) from 0 to 10
ft/sec from one point of reference, but accelerates from 10 to 20 ft/sec
from another point of reference. People at the points of reference would
come up with different measurements for how much energy the object gained in
its acceleration (supposedly). That is very strange.
2. Secondly, suppose from one point of reference in space an object
accelerates from 0 to 1 ft/sec using E amount of energy. Now suppose the
object from another point of reference accelerates again from 0 to 1 ft/sec,
or 1 to 2 ft/sec from the other point of reference, using again E amount of
energy. The ship used the same amount of energy to increase from 1 to 2
ft/sec as it did to increase from 0 to 1 ft/sec, but using KE=.5mv^2 it
should have used 3 times as much energy.
It just happens (?) that the W=Ft equation solves these two problems
with the KE equation. How? The energy equation can be derived from the work
equation. The energy equation comes out as E=mv (the same as momentum).
Newton's equation for energy (KE=mv) I feel was correct.
Send email to JoelM...@USA.net if you think you can prove me wrong or
right.
Joel Merritt
Johnathan Smith
Joel Merritt wrote:
Work and EnergyWork is closely associated with energy, because work can be defined as a
change in kinetic energy (KE) or movement of a mass. Sir Isaac Newton told
us that work equals FD where D is the distance an object is moved directly
against a force of F. Newton also believed that the amount of motion (KE) in
a moving mass is MV where the mass is moving at velocity V. Newton cannot
have been correct in both of these equations, because these two equations
(W=Fd and KE=mv) contradict one another. Newton's work equation was accept
as correct while the energy equation was rejected. I believe that Newton's
work equation is incorrect and the KE equation was correct. Now let me
explain why I believe this.
Both the work equation and the KE equation bothered me, because some
things didn't add up about them.1. Suppose a space ship is accelerating through space at a rate of 10
ft/sec/sec. The ship would be using nearly a constant amount of fuel, but
would be doing much more and much more work (supposedly). The work being
done by the ship would be increasing directly with the velocity of the ship
which varies directly with time.
Not exactly. The work increases with distance. W=Fs where s is the distanceIf
the mass and the rate of acceleration are constant then then the work is
only increased by an increase in distance.
2. Now suppose the same space ship is under the influence of gravity.
The ship is burning the same fuel as before, or using the same amount of
"energy", but if the force of gravity is equal to 10 ft/sec/sec then the
ship would be stationary and (supposedly) doing no work.
So theres no net amount of work. No biggie
3. Suppose a bowling ball is dropped from a skyscraper. The work that
gravity would be doing to the object in accelerating it would vary directly
with time (supposedly). That means that gravity does more and more work for
every unit of time as an object falls .
No problem here
4. The power of gravity would also be increasing.
Force of gravity increases inversly proportional to square of distance...so
All four of these past problems
The only "problem" is the term work isn't really meant to deal
with potential energy which really isn't a problem.
can be solved by defining work as force
times time (W=Ft)
Lets break that down W=force x time
=mass x acceleration x time
=mass x change in velocity divided by change of time x time
Do I need to point it out
rather than force times distance. If this different
equation is not correct then it is a very coincidental thing for it to solve
all of these difficulties. First of all, there shouldn't be many, if any
difficulties with the correct equation. Second of all, there should be many
difficulties with an incorrect equation (such as W=Ft if it is incorrect).
When I came up with the W=Ft equation, I had no idea it would solve the
bunch of problems with the W=Fd equation. It was only because of one of the
problems that I thought W=Ft made more sense (it was quite a coincidence [?]
when it solved my other logic problems).
Kinetic Energy (KE) is today defined as .5mv^2 (one half mass times
velocity squared), but is that correct? I will give a few reasons why I
question it.1. Suppose an object accelerates (in a Galilean universe) from 0 to 10
ft/sec from one point of reference, but accelerates from 10 to 20 ft/sec
from another point of reference. People at the points of reference would
come up with different measurements for how much energy the object gained in
its acceleration (supposedly). That is very strange.
No they wouldn't. In each frame the object whose mass isconstant only gains
10 ft/sec of velocity. General Relativity aside
kinetic energy is relative but changes in kinetic energy are not.
2. Secondly, suppose from one point of reference in space an object
accelerates from 0 to 1 ft/sec using E amount of energy. Now suppose the
object from another point of reference accelerates again from 0 to 1 ft/sec,
or 1 to 2 ft/sec from the other point of reference, using again E amount of
energy. The ship used the same amount of energy to increase from 1 to 2
ft/sec as it did to increase from 0 to 1 ft/sec, but using KE=.5mv^2 it
should have used 3 times as much energy.
Same as above
It just happens (?) that the W=Ft equation solves these two problems
with the KE equation. How? The energy equation can be derived from the work
equation. The energy equation comes out as E=mv (the same as momentum).
Newton's equation for energy (KE=mv) I feel was correct.
Same as above
Any1
From Newton: F = m*a = m*dv/dt => F*dt = m*dv
Integrate to obtain F*t = m*v : Just what you had, but it's called MOMENTUM.
Let's start again:
F = m*a = m*dv/dt = m*(dv/dx)*(dx/dt) , in which (dx/dt) is of course equal to
v.
F = m*v*dv/dx => F*dx = m*v*dv
Integrate to obtain F*s = 0.5*m*v^2
You see, they both follow from newton's law. One does not preclude the other.
They're just different ways to note it down. Work is defined to be F*s. To
illustrate the downside
of your definition, consider how much work is done by gravity if a bowling ball
(with mass m) falls from a skyscraper with heigth h.
First, here is the W = F*s method: W = (m*g)*h = mgh.
Now, W = F*t : W = (m*g)*t ; t = h/<v> ; <v> = 0.5*a*t = 0.5*g*t with <v> the
mean speed.
t = h/(0.5*g*t) => t^2 = 2h/g => t = (2h/g)^0.5 => W = m*(2hg)^1/2 .
Which one is easier?
Now the amount of work done when the ball falls for t seconds.
W = F*t : W = m*g*t = mgt
W = F*d = 0.5 mv^2 : F*t = m*g*t = m*v => v = g*t => W = 0.5*m*(gt)^2
Which one is easier?
Yes, F*t, but it's not that much easier and would you want it in the first
situation?
Any1
Joel Merritt wrote:
> Work and Energy
>
> Work is closely associated with energy, because work can be defined as a
> change in kinetic energy (KE) or movement of a mass. Sir Isaac Newton told
> us that work equals FD where D is the distance an object is moved directly
> against a force of F. Newton also believed that the amount of motion (KE) in
> a moving mass is MV where the mass is moving at velocity V. Newton cannot
> have been correct in both of these equations, because these two equations
> (W=Fd and KE=mv) contradict one another. Newton's work equation was accept
> as correct while the energy equation was rejected. I believe that Newton's
> work equation is incorrect and the KE equation was correct. Now let me
> explain why I believe this.
> Both the work equation and the KE equation bothered me, because some
> things didn't add up about them.
>
> 1. Suppose a space ship is accelerating through space at a rate of 10
> ft/sec/sec. The ship would be using nearly a constant amount of fuel, but
> would be doing much more and much more work (supposedly). The work being
> done by the ship would be increasing directly with the velocity of the ship
> which varies directly with time.
> 2. Now suppose the same space ship is under the influence of gravity.
> The ship is burning the same fuel as before, or using the same amount of
> "energy", but if the force of gravity is equal to 10 ft/sec/sec then the
> ship would be stationary and (supposedly) doing no work.
> 3. Suppose a bowling ball is dropped from a skyscraper. The work that
> gravity would be doing to the object in accelerating it would vary directly
> with time (supposedly). That means that gravity does more and more work for
> every unit of time as an object falls .
> 4. The power of gravity would also be increasing.
> All four of these past problems can be solved by defining work as force
> times time (W=Ft) rather than force times distance. If this different
> equation is not correct then it is a very coincidental thing for it to solve
> all of these difficulties. First of all, there shouldn't be many, if any
> difficulties with the correct equation. Second of all, there should be many
> difficulties with an incorrect equation (such as W=Ft if it is incorrect).
> When I came up with the W=Ft equation, I had no idea it would solve the
> bunch of problems with the W=Fd equation. It was only because of one of the
> problems that I thought W=Ft made more sense (it was quite a coincidence [?]
> when it solved my other logic problems).
> Kinetic Energy (KE) is today defined as .5mv^2 (one half mass times
> velocity squared), but is that correct? I will give a few reasons why I
> question it.
>
> 1. Suppose an object accelerates (in a Galilean universe) from 0 to 10
> ft/sec from one point of reference, but accelerates from 10 to 20 ft/sec
> from another point of reference. People at the points of reference would
> come up with different measurements for how much energy the object gained in
> its acceleration (supposedly). That is very strange.
> 2. Secondly, suppose from one point of reference in space an object
> accelerates from 0 to 1 ft/sec using E amount of energy. Now suppose the
> object from another point of reference accelerates again from 0 to 1 ft/sec,
> or 1 to 2 ft/sec from the other point of reference, using again E amount of
> energy. The ship used the same amount of energy to increase from 1 to 2
> ft/sec as it did to increase from 0 to 1 ft/sec, but using KE=.5mv^2 it
> should have used 3 times as much energy.
>
> It just happens (?) that the W=Ft equation solves these two problems
> with the KE equation. How? The energy equation can be derived from the work
> equation. The energy equation comes out as E=mv (the same as momentum).
> Newton's equation for energy (KE=mv) I feel was correct.
>