A 3 kg mass is moving along a frictioless surface at 17 m/s. It then reaches a slope and starts up it. The slope has an angle of 10º and a small amount of friction. What is the coefficient of friction on the slope if the block slides 4.5 m up the slope before stopping? m=3 kg v(i)=17 m/s v(f)=0 m/s d=4.5 m a=v/d*v F(g)=mg F(N)=F(g)cos(theta) F(kf)=F(g)sin(theta) F(kf)=(mu)(kf)F(N) a=64.2m/s^2 F(g)=29.4N F(N)=29.0N F(kf)=5.1N (mu)(kf)=F(kf)/F(N) =5.1N/29.0 =0.18??? I can't seem to determine the F(net) equation. I also can't figure out why V(i) and d would be provided if they aren't a necessary part of the F(kf)=(mu)(kf)F(N). Any help making sense of this problem would be great.