>>14346146How do you define the spring stiffness constant in the first place?
I was taught it is the spring force per elongation i.e. $k = F_s / \Delta L$
So obviously $k=mg/\Delta L$
The other solution you mentioned
>mg*?L = 1/2*k*?L^2>This is setting the work done by potential energy and spring potential equal to each other, but does it apply for the stiffness constant?Well, if you think about it, work is the integral of force along a path. In this case ?L. You can easily see this as the force for gravity is mg and spring stiffness is k?L. Integrating these forces over ?L gives makes mg?L and 1/2k?L^2.
So you're not comparing the same thing.
As
>>14346432 and
>>14347080Pointed out, you're looking at different aspects of equilibrium.
When looking at force balance, you're determining where the mass undergoes has zero net force acting on it, so where the acceleration is zero and where the mass would come to rest over time as all kinetic energy has been dissipated.
When looking at the work balance, you find what displacement corresponds to changing all the height potential energy into spring elongation potential energy, so what the elongation would be when all height energy is transferred to strain energy. This can come in handy, but it is not how the spring stiffness constant is defined.
