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?(adsbygoogle = window.adsbygoogle || []).push({});

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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# Forces on a slope

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