Physics
posted by lizgreene52 .
Sukey is riding her bicycle when she sees a fallen tree 42 m ahead. If the coefficient of friction is .36 and she is travelling at 50 km/ hr and she and the bike have a mass of 95 kg, how much stopping distance does she need. I got as far as her speed equals 13.9 m/sec. Do we need her Normal force (95 x 9.8m/sec 2) = 930N.

10c<0.06>=f(n)(l1/l3+l2/l4)
uf=f(x)_ln<1.962>
fx/%fl
%fl=1.333
2C<0.09>=f(1.33)
d=? 
Sorry, I don't understand all the abbreviations.

Sorry, I don't understand all the abbreviations.
I figure that her stopping speed, 0 = 13.9m/sec + 2ad where a = F/m but here I get stuck. I don't know exactly what to use for F
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