Tell Me About Kinetic Energy, Please

[dock]

Tell Me About Kinetic Energy, Please!

according to the traditional physics the kinetic energy is:

E=mVV/2;

let that speed be constant and nonzero => E<>0;

according to Newton the 1st whenevr the speed V=const the resulting force acting upon that object is zero;

the energy by definition is:
E=FxX;

since F=0 => E=0

HOW ABOUT THAT?

 

[FZ+]

Erm... no?

Work done = change in energy is the integral of force with respect to distance.

All this shows is that there is no change in kinetic energy for an object moving at constant velocity.
Which is obvious, really.

 

[FZ+]

And use * instead of x for the multiplication operator. It's less confusing that way.

 

[dock]

Originally posted by FZ+
Erm... no?

Work done = change in energy is the integral of force with respect to distance.

All this shows is that there is no change in kinetic energy for an object moving at constant velocity.
Which is obvious, really.

i'm not talking about dE but E.
OK so dE=FdX then integrated we have (E=FX and F=const)
d(E=FX) <=> dE=FdX cause F=0=const.
so it's not only dE=0 but E=0 also!

when F=const then dE=FdX
when E=const then FdX=-XdF
when X=const then dE=XdF
when none then
dE=FdX+XdF

supportive members, please, be louder!

 

[FZ+]

Let me repeat:

YOU HAVE SHOWN THE CHANGE IN ENERGY.

YOU HAVE SHOWN THERE IS NO CHANGE IN ENERGY.

F * x = Work Done = change in energy

Integral of F wrt x gives F*x + C

The constant of integration C is the initial kinetic energy of the system.

UNDERSTOOD?

 

[dock]

Originally posted by FZ+
Let me repeat:

YOU HAVE SHOWN THE CHANGE IN ENERGY.

YOU HAVE SHOWN THERE IS NO CHANGE IN ENERGY.

F * x = Work Done = change in energy

Integral of F wrt x gives F*x + C

The constant of integration C is the initial kinetic energy of the system.

UNDERSTOOD?

F * x = Work done?
ain't Work done = dE?
how come only one derivate in that equation.that violates the mightiest of the laws:"Every change is simultaneous with at least one other".
& int Fdx = F(x(2)-x(1))
but it's also
& int dE = E(2)-E(1)
and then
E(2)-F(2)x(2)=E(1)-F(1)x(1)=const=0 threrefore

E=Fx

cause

F(1)=F(2)=F
you cannot put starting condition on the staring conditions.it leads no where that's whay that const=0.

 

[russ_watters]

Its just different notations in different equations, dock. Apparently wherever you got that from uses dE and E interchangeably. And its not unreasonable to do so since often times a change in E results in a conversion from one form of E to another (such as kinetic being converted to potential).

 

[STAii]

dock, from the Work-Energy theorom, the change in energy in a system is equal to the work.
So F*x (the work) is equal to dE in the system.

E(2)-F(2)x(2)=E(1)-F(1)x(1)=const=0 threrefore

It is right that when F=0, E=const, but why are you taking this const to be 0 ?

 

[dr-dock]

this kinetic energy paradox aplies to the photon as well.
the photon has constant speed c => the force is zero => the energy of the photon is zero cause E=FxX=0xX=0.