Posted by Anonymous on .
1)A logging truck is dragging a 427 kg log through the forest at a constant speed of 3.5 m/s. If the truck is applying a force of 1047 N to the log to keep it moving, what is the coefficient of friction between the log and the ground?
im thinking you use the kinetic friction= μk(normal force) equation so does that mean it would be 3.5 m/s=μ(1047N) but then where does the 427 kg come in??

Phyics 
drwls,
The speed has nothing to do with it. You just need to know that it is constant. That means the net force is zero.
Towing force = Friction force
1047 N = M*g*μk
= 427*9.8*μk
Solve for μk