work energy theorem and minimum potential energy theorem.

Dear Sir,
In case of a spring mass system,
The energy stored in the system after the displacement x is
=1/2 *k*x^2
The work done by the force F on the system is
= F *x
According to work energy principle both should be equal
and hence
1/2 k*x^2 = F *x
But according to minimum potential energy theorem , they are not treated as equal. i.e The potential energy of the system is taken as P.E = 1/2 k*x^2 - F *x
and it should be minimum.
What is the relation between both the principles? One treats both the force are equal and other treats there is a difference between them.
I am confused. regds, Yogesh Joshi
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yogesh wrote:

Right.
Only for a constant force F! In General: W = \int F(x)*dx

Wrong, since F(x) = k*x is not constant.

The "minimum potential energy theorem" is used to find the equilibrium. Here we have potential Energy U = 1/2 k*x. Therefore we find the equilibrium to be at x = 0.
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snipped-for-privacy@indiatimes.com (yogesh) wrote in message

That part is right.

Sure, assuming the spring's compression energy is zero at x = 0.

No, not quite. F here is a function of x, namely F = - k x. The -ve sign is an indicator that this is a restoring force, that is, the force is trying to bring the spring back to x=0. This means the energy is lowest at x = 0. And also, it's an integral.
W = integral_0 ^X F(x) x dx = - 1/2 k X^2
And with a little work at getting the signs right, and realizing how the compression energy of the spring is related to the energy of a mass on the end of the spring, you will see that things work out. Socks
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snipped-for-privacy@indiatimes.com (yogesh) wrote in message

There's your problem. If the force you apply is constant, the the equation W=F*x is correct. HOWEVER, for a spring system, the force you apply is not constant, and is changing with time. The force applied to a spring system is:
W= integral ( F(x)*dx )
where F(x)=k*x. Therefore, you have:
W= integral ( F(x)*dx ) = integral ( k*x*dx ) = .5kx^2
Hope that helps! Dave
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yogesh wrote:

Indeed. Try plotting the force as a function of x. Don't forget the contribution that is accelerating m.
Cheers
Greg Locock
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