Why the work done on the spring is different?

Question

If a Woman weighing $600 N$ steps on a bathroom scale containing a stiff spring, the spring is compressed $1.0 cm$ under her weight, the spring constant is $60000 N/m$.It turns out that the work done on the spring is $\frac{1}{2}kx^2=\frac{1}{2}(60000N/m)((0.01)(0.01))=3$ Joules. BUT isn't the work done on the spring $600N$ times $0.01m=6$ joules? Since the woman exerts a constant $600N$ force on the spring throughout the $0.01m$ displacement? Thanks in advance.

Answer 1

You are correct in saying that the work done by gravity is $6\ \rm J$. So where did the energy go?

Well consider the case where the woman is at rest above the uncompressed spring (magically hovering, I don't know) and then is released. Using energy conservation we find that $$\frac12kx^2=mgx$$ or $$x=\frac{2mg}{k}=2\ \rm{cm}$$

So we see that, if we had an ideal system of releasing the woman from rest above the spring she would actually sink $2\ \rm{cm}$ before coming to rest for the first time. But in your problem she only goes down by half this amount.

We must conclude that energy is lost due to dissipative forces. This makes sense. Without dissipative forces the woman would oscillate on the scale indefinitely. The dissipative forces are needed for the woman to come to a complete, lasting stop.

Therefore, gravity does $6\ \rm J$ of work. Half of it goes into the spring potential energy. The other half leaves the system through whatever mechanisms bring the woman to rest after stepping on the scale. (And of course this is all just focusing on if she starts right above the uncompressed spring at rest)

Answer 2

The work provided is 6 J. But only part of this work is converted into elastic potential energy. The rest is converted into kinetic energy (the mass oscillates) and finally in heat. You would need a quasistatic transformation to convert all the work into elastic potential energy.