1. The problem statement, all variables and given/known data A grindstone in the shape of a solid disk with diameter .52m and mass 50 kg is rotating at 850 rev/min. You press an ax against the rim with a normal force of 160 N and the grindstone comes to a rest in 7.5 seconds. Find the coefficient of kinetic friction of the ax on the grindstone. 3. The attempt at a solution Since the friction causes the grindstone to come to a stop, then the initial kinetic energy of the grindstone must equal the negative of the work done by friction. And the force of friction is given by the coefficient of friction, k , times the normal force, n, so the work done will be that force times the distance x. Using the work energy theorem for the grindstone we would get W=-.5Iv^2, where I is the moment of inertia and v is its initial speed. I=MR^2/2=(50*.26^2)/2=1.69 kgm^2 v=850 rev/min=89rad/sec x=89(7.5)(2pi.26)=1090 m knx=-.5Iv^2 k=(.5Iv^2)/nx k=(.5)(1.69)(89^2)/(160)(1090)=.038 This is obviously wrong and it doesn't match the answer of.482 given by my text, but where am I going wrong? Thanks for the help.