How many $g$'s does a mass see as it impacts and compresses a spring? Essentially, I'm designing a drop test, and to simplify my problem I've modeled the system as a known mass $m$ being dropped from a set height $h$ onto a spring (set $k$), compressing it ($x$). How many $g$'s does the mass see upon impact, when the spring is fully compressed?
Using energy equations to solve for $x$ is simple enough (there are even online examples like this YouTube video)
Once I've solved for $x$, does it make sense to use $a = kx/m$ --> $g$'s $= a/g = a / 9.81m/s^2$?
I have one more question: If using the method above is reasonable, it seems that the smaller the mass, the greater the acceleration and the greater the $g$'s; how is it possible? I would have thought that the smaller the mass, the less $g$'s are felt on impact.
 A: 
Once I've solved for $x$, does it make sense to use $a = kx/m$ --> $g$'s $= a/g = a / 9.81m/s^2$?

Yes, according to the Wikipedia definition of g-force, it is common to express an acceleration in terms of $g$. I.e. an acceleration of $98.1\,\text{ms}^{-2}$ can be said to be an acceleration of "10 g". Your equation translates this correctly by dividing the acceleration by $g$.

If using the method above is reasonable, it seems that the smaller the mass, the greater the acceleration and the greater the $g$'s; how is it possible? I would have thought that the smaller the mass, the less $g$'s are felt on impact.

No, your intuition is wrong - a smaller mass does experience a larger acceleration, but not simply because of the equation $a=\frac{kx}{m}$.
If you imagine being in a very heavy metal ship, dropped onto a large spring, you are going to not feel that much acceleration as there is a larger inertia to accelerate, and although there is a larger force from the spring, it is not that much larger than for someone in a small car.
Maybe it is easier to see from the maths...
The GPE is converted to elastic potential:
$$mgh = \frac{1}{2}kx^2$$
This means $x \propto \sqrt m$.
Your equation say that $a \propto \frac{x}{m} \propto \frac{\sqrt m}{m} \propto \frac{1}{\sqrt{m}}$, so as the mass increases, the max acceleration decreases.
