How can I apply a certain "pressure" (g/cm²) to the ground by dropping a weight? I'm trying to find a way to apply 100g/cm² to the ground (substrate = snow) by dropping a weight from a particular height.
If I drop 250g weight (80cm² area) from 1m, I'll get a velocity of 4.43m/s² and kinetic energy of 2.45 Joules... or 24983 gram force*cm. I could figure out the impact force by assuming a 0.1s impact, but I can't really see that this will help me...?
If i know the area of the weight is 80cm² how can I figure out what the "pressure" is over that area (ultimately get to g/cm²) given that I know the work (gf*cm)?  
I think I'm just missing some basic physics principles in trying to think through this - I'd appreciate the help (and thanks).
 A: Here is a bit of basic physics.
If you drop a mass $m$ with area $A$ from a height $h$ onto snow, and it penetrates the snow to a depth $d$, then the average pressure on the snow during the fall is calculated as follows:
Total distance dropped: $D = h+d$.
Total gravitational energy: $E = mg(h+d)$ 
Retarding force $F$ acted over distance $d$ to do the same amount of work as gravity:
$$F d = mg(h+d)$$
The force per unit area (average pressure) is then given by:
$$P=\frac{F}{A}=\frac{mg(h+d)}{Ad}$$
Interestingly, the term in the denominator $Ad$ is the volume of snow that is compacted, while the term in the numerator is the energy (weight times total height dropped).
Obviously, there are a few things to think about when you do this with "real" snow: in particular, the force will probably change with distance, but as long as you are falling about the same distance (same depth as an actual wolf's footprint) you will have roughly the same physics. Also, as a wolf runs it will may at times have most of its weight on just one foot, not four - although if the snow is soft, it may change to a trot (opposite diagonals), to spread its weight more evenly.
