Question

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

Answer 1

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.