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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

Reply to
yogesh
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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*x². Therefore we find the equilibrium to be at x = 0.

Reply to
Theo Wollenleben

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

Reply to
puppet_sock

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

Reply to
David Harper

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

Cheers

Greg Locock

Reply to
Greg Locock

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