Find Coefficient of Friction on Slope w/ 3kg Mass Moving at 17 m/s

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A 3 kg mass moving at 17 m/s reaches a 10º slope with friction and slides 4.5 m before stopping. The calculations involve determining the acceleration, gravitational force, normal force, and kinetic friction force. The coefficient of friction is calculated to be approximately 0.18. The initial velocity and distance are necessary for applying kinematic equations to find the net force. Understanding these principles is crucial for solving similar physics problems effectively.
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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.
 
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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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