Solving the Question of Units for Dropped Mass on Surface

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I *think* this is right, but if there is something obviously wrong please tell me where


Ive got a mass of weight 'y' dropped onto a surface that can support 10y
there is no wind resistance
the mass starts from stationary and accelerates under gravity (assumed to be 9.8 m/s/s)
the distance is 3m

constant acceleration from stationary for s distance is v^2=2as. v^2=2(9.8)3=7.6 m/s
The kinetic energy of a particle of mass m moving at velocity v is 1/2*m*v^2 K=1/2(y)(7.6^2)=29.4y

so the particle would excede the resistance by 19.4y

with respects to the different units involved is this correct, or can i not take the resistance of the surface (measured in neutons) away from the kinetic energy of the particle (measured in joules) to get an answer because of the different units?
 

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The kinetic energy of a particle of mass m moving at velocity v is 1/2*m*v^2 K=1/2(y)(7.6^2)=29.4y
The relationship between mass m and weight y is m = y/g.

Energy/work is equated to force * distance, so 1 Joule (J) = 1 Newton * 1 m = 1 N-m.


Force accelerates/decelerates a mass.

It would seem one needs to determine the deflection of the surface or the time needed to decelerate the mass y/g in order to determine the force.

http://hyperphysics.phy-astr.gsu.edu/hbase/impulse.html