Elastic Deformation through compression (stress)

[fanie1031]

1. The total cross-sectional area of the load-bearing calcified portion of the two forearms bones (radius and ulna) is approximately 2.16cm^2. During a car crash, the forearm is slammed against the dashboard. Tha arm comes to rest from an initial speed of 81.9km/h in 7.72 s. If the arm has an effective mass of 1.88kg what is the compressional stress on the arm? Answer in units of Pa.
2. Stress=F/A ,
additional
F=ma
Vf=Vo+at
3.
Conversions
A=2.16cm^2*1m^2/10^4cm
=2.16e-4m^2
Vo=81.9km/h*(1hr/60mins)(1min/60s)(1000m/1km)
=22.75m/s

F=ma
a=(Vf-Vo)t=-2.94689.

I assumed I could make it positive since the problem's context is in forward motion not retardation.

F=1.88kg*2.94689
=5.540kg*m/s^2

P=F/A
P=5.540kg*m/s^2/2.16e-4
=25.648.8677N/m^2= same in Pa.I feel like I put in the appropriate conversions and formulas but when I plug in the answer, I'm told I am wrong. Does it have to do with the fact that it mentions "two forearm bones" in the beginning? If it is, where do I half the cross sectional area?

I hope I'm not too far off.

 

[LowlyPion]

Your method looks fine. Units are OK.

Check the significant digits is about all I can suggest.

Your concern about 2 cross sections should be allayed by the statement of total cross section.