Posted by lizgreene52 on .
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.

Physics 
PhysicsPro,
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=? 
Physics 
lizgreene52,
Sorry, I don't understand all the abbreviations.

Physics 
lizgreene52,
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