A 2.30-kg block starts from rest at the top of a 30.0° incli

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A 2.30-kg block starts from rest at the top of a 30.0° incline and slides a distance of 1.90 m down the incline in 1.80 s.
(a) Find the magnitude of the acceleration of the block.
_________ m/s2
(b) Find the coefficient of kinetic friction between block and plane.
_____________
(c) Find the friction force acting on the block.
magnitude
_________ N
____________direction ( up the incline,down the incline,normal to the incline and upward,normal to the incline and downward)
(d) Find the speed of the block after it has slid 1.90 m.
_____________ m/s

a) x = v0t + 0.5at^2； v0 = 0
So a = 2x/(t^2) = 2*1.9/(1.8^2) = 1.17ms^-2
b) Net force on the block = ma = mgsin(30) - μmgcos(30)
so gsin(30) - μgcos(30) = a
(0.5)(9.8) - 9.8μcos30 = 1.17
9.8μcos30 = 3.73
9.8μ = 4.31
μ = 0.440
c) Frictional force = μN = μmgcos(30) = 0.440*2.30*9.8*cos30 = 8.58N up the incline (since the block is moving down the incline,Ff is in the opposite direction)
d) v = v0 + at = 0 + 1.17*1.8 = 2.11
I think; I haven't done mechanics stuff in forever.

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