ddr

Ok,

an object travels with V=const and it caries E=mVV energy.
But, according to Newton I if V=const then F=0 or no force acts upon it.
On the other hand E=FD is energy for distance D and it's zero if F=0.
So the const velocity traveling object both has nonzero and zero energy!?!?!?!?!?!
How come?
What about photons traveling at c=const?

Who ever solves this paradox, coffe is on me?!?

F*D is not energy , it's work

ddr
I'm sorry but...

E=FD is energy
dE=FdD+DdF is work done.

I dont think you understand Newton's laws of motion. The best way to describe this is with a free body diagram.
___________Fg
____________|
____________v
Fapp---->Object<----Fopp
____________^
____________|
___________Fn

(please excuse the underscores, the formatting of the forum doesnt seem to allow for ASCII art)

Fapp is the applied force, Fopp is the force opposing the applied force (Whether it's friction, wind resistance, or whatever), Fg is the force of gravity on the object, and Fn is the normal force exerted by the ground on the object (due to the 'equal and opposite reaction' law of Newton's)

Now as long as Fapp and Fopp are the same, there will be zero net force, which is what you're talking about. This means there is zero work done TO the object by some other object. W=FappD where D is in the direction of the Fapp. If Fapp=0, then W=0 as well. When something is moving at a constant speed it means that it's impossible to discern if the object is the thing moving, or if it is stopped and everything around it is.

In the case of photons, however, they have zero rest mass. This means that from the moment they were created, all the energy that made them was for the purpose of moving at light speed. In that sense, they have entirely kinetic energy, and no potential energy.

And what the hell is that second equation you gave? It looks like a bunch of capital Ds mixed with some lowercase Ds with a few Es and Fs thrown in there for good measure. Please explain what each of the variables are because it certainly doesnt look like anything I've seen representing 'work done' ever before.

Janus
Staff Emeritus
Gold Member

Originally posted by ddr
Ok,

an object travels with V=const and it caries E=mVV energy.
But, according to Newton I if V=const then F=0 or no force acts upon it.
On the other hand E=FD is energy for distance D and it's zero if F=0.
So the const velocity traveling object both has nonzero and zero energy!?!?!?!?!?!
How come?
What about photons traveling at c=const?

Who ever solves this paradox, coffe is on me?!?

E = mv²/2
E = fd

if you take mv² and break it down into base units you get

E = md²/t²

if you take fd and break it down into
base units:

first f=ma so

a= d/t² so

E= m(d/t²)d

which rearranges to

E = md²/t²

fd just tells you how much energy is imparted to an object if you exert a force on it over a given distance.

mv²/2 tells you how much energy the object has is moving at a given v

For instance, if exert a given force on a object it will accelerate. if I continue to exert that force while the obect moves over a given distance, it will have both gained a given velocity and attained a given amount of energy.

If I quit exerting the force now, the energy gained is still there, as is the velocity, and I gain relate the amount of energy to the velocity.

ddr
the paradox is not within the counter parts incompatibility.Actuaslly they match perfectly.it's only that from one point of energy view that object has nonzero energy while from another it has zero energy at the same time.

you wanna know what i think?

i think that force is proportional with displacement ie. displacement takes place in the same direction of the actual force. so it's not velocity that has same direction with displacement in general. it's covered with this equations:
FdD>0 and if F=0 then dD=0;
Newtonian kinetic energy E=mVV is pure wind in the fog.
when force matters there is only one way to estimate energy E=FD where D is the equilibrium distance.
Newton the 1st should be if dD<>0 then F<>0 and if F=0 then dD=0 ie when there is no force there is no displacement.
the confusion is a pure result of confusious traditional kinetics.

I'll revolutionize physics one way/day or another.

You can't revolutionize physics with that idea. It's called work...

W=Fd

Work = Force applied * Distance over which the force was applied.

This is not a new idea. The amount of work done to an object is the amount of energy one adds to the object to make it move a certain distance. That means your E=fd is nothing new.

Sorry

ddr
Originally posted by cytokinesis
You can't revolutionize physics with that idea. It's called work...

W=Fd

Work = Force applied * Distance over which the force was applied.

This is not a new idea. The amount of work done to an object is the amount of energy one adds to the object to make it move a certain distance. That means your E=fd is nothing new.

Sorry
member, hear me now!!

dE=W=FdD+DdF.

work is not only the product of the force and displacement but it's also consisted of distance/position times disforcement.if one object remains in same position and its force changes then its energy changes as well. the fact that the object is doing work tells me that it contains energy in every event during changes...

E=FD is not new? true. what is new is its interpretation.

the main problem with the physicists is that they get lost while interpreting the simpliest math...

Hurkyl
Staff Emeritus
Gold Member
if one object remains in same position and its force changes then its energy changes as well.

Recall that an object can remain stationary (or with a constant velocity) only if the net force is zero (otherwise there would be a net acceleration).

Originally posted by ddr
member, hear me now!!

dE=W=FdD+DdF.

work is not only the product of the force and displacement but it's also consisted of distance/position times disforcement.if one object remains in same position and its force changes then its energy changes as well. the fact that the object is doing work tells me that it contains energy in every event during changes...

E=FD is not new? true. what is new is its interpretation.

the main problem with the physicists is that they get lost while interpreting the simpliest math...

It usually helps when you explain what all the variables in your equation represent. Technically, (if i'm reading this right) you're saying sort of the equivalent to W=2(Fd). Work can't equal Fd along with some extra variables. If you're going to make a new interpretation of how work changes, you need to change the formula so that work comes out being the same in both cases.

ddr
Originally posted by Hurkyl
Recall that an object can remain stationary (or with a constant velocity) only if the net force is zero (otherwise there would be a net acceleration).
yeah Hurkyl,
my statement got out of context a bit.i meant just like the force is causing displacement, the position/distance is causing disforcement as well.

You wanna know how come I know?

Here it is:
assume one object is in equilibrium state and you wanna get it out of there.what you need to do is invest some force while the equilibrium distance is zero.The object will shift where the force points.the further away the object gets, the smaller the force becomes.so at some extreme point of the drive the force will become zero.then the force is disforcing in the oposite direction of the equilibrium distance.

so we have to events:
starting one (force, distance)=(Fmax, 0)
ending one (force, distance)=(0, Dmax)
the total energy will be E=Fmax*Dmax

here is how the drive works:
Fnow=Fmax*cos(alpha)+Dmax*sin(alpha)*|Fmax|/|Dmax|
Dnow=Dmax*cos(alpha)+Fmax*sin(alpha)*|Dmax|/|Fmax|
E=const (if the oscillation is harmonic)
you can notice that Fnow*Dnow=Fmax*Dmax=E=const then

You're assuming that the force is gradually decreasing. Thats not how it works. according to newton, an object in motion stays in motion and an object at rest stays at rest unless acted upon by an outside force. A single push on a box on a frictionless surface, keeps it going infinitely. The work done is minute because the distance over which the force is applied is minute. the d in W=Fd is not the distance over which the object travels, it's the distance over which the force is applied. After you quit accelerating, no extra force is being added to the car, it's coasting along because of its inertia.

russ_watters
Mentor
Originally posted by ddr
I'll revolutionize physics one way/day or another.
Good luck. You'll need it. W=FD is so basic as to be practically a definition. It won't easily be changed.
the further away the object gets, the smaller the force becomes
That of course depends on the nature of the force. What you are describing sounds like an oscillating spring. Not all forces work that way.

No offense, ddr, but it simply sounds to me like you haven't learned enough about Newtonian Physics.

ddr
Originally posted by cytokinesis
Thats not how it works. according to newton, an object in motion stays in motion and an object at rest stays at rest unless acted upon by an outside force. A single push on a box on a frictionless surface, keeps it going infinitely. The work done is minute because the distance over which the force is applied is minute. the d in W=Fd is not the distance over which the object travels, it's the distance over which the force is applied. After you quit accelerating, no extra force is being added to the car, it's coasting along because of its inertia.
who said newton was right?>
-where the force is acting there the displacement is taking place.No force no displacement.ain't that obvious.force cannot be related with velocity because this way it is related with time while force cannot depend on time cause time is not matter quality while force describes matter interaction within every instance of time and therefore force can only be function of matter qualities like charge,mass or position(distance).

Integrate F(r(t)) dot r(t) from t_{bigbang} to t_{present} for any particle in a conservative field, behold the kinetic energy is precisely the value reckoned.

Your confusion arises when you try to start integrating at some time after F(r(t)) goes down to 0, ignoring the previous eternity in which the particle's incipient energy was imbued. Falling into the indigenous guile of employing such finitely applicable results as W = Fd and E = 1/2mv^2 when you have disrupted the fragile circumstances in which associations between the two are valid, in the system put forth, your theoretics yield nidorous quandary.

Last edited:
russ_watters
Mentor
Originally posted by ddr
who said newton was right?
I'll say it: Newton is right. You are not. You should learn more about Newton's theories and how they actualy work (and they do actually work).
where the force is acting there the displacement is taking place.No force no displacement.ain't that obvious.
Have you ever stood on a bathroom scale...? Clearly the scale measures a force that is constant for as long as you stand on it. And yet there is no displacement.

Integral
Staff Emeritus
Gold Member
Hey Doc, how ya' doing. Been awhile! Looks like you've read a bit more, learned a bit but still want everybody else to be wrong, and you right. Good luck. I am not even going to try an explain it to you.

If others don't remember Doc, he is an old timer, do a search on PF2 archives to get a history.

ddr
Originally posted by Integral
Hey Doc, how ya' doing. Been awhile! Looks like you've read a bit more, learned a bit but still want everybody else to be wrong, and you right. Good luck. I am not even going to try an explain it to you.

If others don't remember Doc, he is an old timer, do a search on PF2 archives to get a history.
Integral, amigo, long time no reply to my posts from you. You are corect it's me the well known member dock but not known by good things.A volf might dress him up in karaoke dress but it still goes after the sheeps.I have special intention to get you involved cause you are not just anybody but the one with authority. They listen to you just like I don't.Here is a double-dare for you int.:
This is equation about action Ea=Fa*Da
This is equation about reaction Er=Fr*Dr
E=energy
F=force
D=equilibrium distance.
hint: imagine all that in a "LEVER"
Now Fa=(Dr/Da)*(Ea/Er)*Fr means that newton's 3rd law is invalid when (Dr/Da)*(Ea/Er)<>-1, RIGHT?

But what the heck I'm only a barbarian?
I won't post anything else for a certain while.I'm just tired of it all.They just kicked me of http://physlink.com
By the way this ain't no productive way to achive something.

neutroncount
Glad you're gone. Stop making up equations like:

This is equation about action Ea=Fa*Da
This is equation about reaction Er=Fr*Dr

and maybe we'd take you seriously.