[Phys-L] Re: Spring question

[titus at MAILAPS.ORG (Aaron Titus)]

You have neglected the work you do on the object in lowering it to its
equilibrium position. If you lower it slowly such that there is no
change in kinetic energy, then you must apply a varying force F_hand
that keeps the object in equilibrium as you lower it. This force goes
to zero at the "equilibrium point" where the force of the spring
balances the gravitational force.

Thus,

\Delta U_grav + \Delta U_spring = W_hand = integral F_hand dy

Here's another way to solve it. Consider that you release it from the
point where x=0 (unstretched spring). When it reaches the equilibrium
position, the object will have kinetic energy of 1/2 mv^2. Thus,

-mgx + 1/2kx^2 + 1/2mv^2 = 0

where v is the speed at the equilibrium point. You can also consider
the kinetic energy term as the "lost energy" after the oscillations
eventually dampen and the object is at rest at the equilibrium
position.

I haven't checked these calculations, by the way. I'm just typing
quickly. You should probably check these for yourself to make sure that
they are correct.

Aaron

On Dec 11, 2004, at 3:16 PM, Savinainen Antti wrote:

> The change in gravitational potential energy of the block
> is mgx. This energy must be in the mass-spring system and
> equals 1/2kx*x (assuming no friction in the spring).
> But this calculation gives m = 1/2* kx/g!!