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.

 

[thomate1]

Your calculations and conversions look correct. It is possible that the discrepancy in your answer could be due to rounding errors. When using conversions, it is important to carry out the full calculation without rounding until the final answer. This ensures that the final answer is as accurate as possible.

Regarding the mention of two forearm bones, it is important to consider the total cross-sectional area of both bones when calculating the stress. In this case, it seems that the 2.16cm^2 already takes into account the total area of both bones. If you are unsure, you can always clarify with the person who provided the problem.

Overall, your approach and calculations seem correct. If you are still unsure about your answer, it would be best to double check with a teacher or another scientist for confirmation.