Forces on a slope

  • Thread starter candycooke
  • Start date
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.
 

Shooting Star

Homework Helper
1,976
4
Upward along the plane, the force ma = -mgsin(theta) - u*mgcos(theta). After the m cancels out, it's a matter of simple kinematics and vi and d are both needed.
 
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