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OpenStudy (anonymous):

A 12.5 kg wooden crate with an initial velocity of 2.5 m/s slides across a rough cement floor for 1.7 m before coming to rest. Find the coefficient of kinetic friction...PLLEEAASSEE! WITH STEPS!!

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OpenStudy (anonymous):

vf2=vi2+2ad -> (2.5)^2=0+2a(1.7) -> 6.25=3.4a -> a=1.83824 Then F - f=ma -> (12.5)(9.8)-kf (12.5)(9.8)=(12.5)1.83824 kf=0.8124

OpenStudy (anonymous):

THHHAAANNNKKKKK YOOOOUUUU SOOOOO MMUUUUCCCCHHHH!!!!!!!!!!!!!!!!!!!

OpenStudy (anonymous):

OH but I just have one question, why in the first part where you used \[vf^2=vi^2+2d\] why do you put the initial velocity as zero and the final velocity as 2.5?

OpenStudy (anonymous):

vf2=vi2+2ad Because the wooden is not moving so the velocity equal -> 0 (First condition vi=0)

OpenStudy (anonymous):

:D ok THANK YOU!

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