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-   -   Derivation of the kinetic energy equation in terms of distance. (http://www.physicsforums.com/showthread.php?t=657794)

CraigH Dec9-12 07:57 AM

Derivation of the kinetic energy equation in terms of distance.
 
I have seen the derivation of the kinetic energy equation using

F=M*v'
and
E=F*x

And I can see how this works, however if you try to do this without thinking about velocity, and only thinking about the rate of change of distance, and the rate of change of rate of change of distance, then the derivation doesn't work, as shown below.

F = Force
M = Mass
x = Distance

Force = mass * acceleration and
Energy = the integral of force with respect to distance:

F = M * x''
E = integral ( F .dx )

sub F into E = integral ( F .dx )

E = integral ( M * x'' .dx )
E = M * integral ( x'' .dx )
E = M * x'

Which isn't true. E should equal 0.5*M*(x')^2
Why does this not work?

Thanks

Doc Al Dec9-12 08:17 AM

Re: Derivation of the kinetic energy equation in terms of distance.
 
Quote:

Quote by CraigH (Post 4190353)
E = integral ( M * x'' .dx )
E = M * integral ( x'' .dx )
E = M * x'

How did you get that last step? (Are mixing up dx with dt?)

CraigH Dec9-12 08:23 AM

Re: Derivation of the kinetic energy equation in terms of distance.
 
Ahhhhh yes I am. Thankyou! I get this now.

CraigH Dec9-12 08:31 AM

Re: Derivation of the kinetic energy equation in terms of distance.
 
In case anyone was wondering

E = M * integral ( x'' .dx )

E= M * integral ( d(dx/dt)/dt .dx)

E = M * integral ( dx/dt .d(dx/dt))

E= M*0.5*(dx/dt)^2

E=0.5*M*(x')^2


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