A 2.10-kg block starts from rest at the top of a 30.0° incline and slides a distance of 2.10 m down the incline in 1.20 s.

Question

(a) Find the magnitude of the acceleration of the block.

(b) Find the coefficient of kinetic friction between block and plane.

(c) Find the friction force acting on the block. magnitide (N) and direction

(d) Find the speed of the block after it has slid 2.10 m.

Answer 1

Alex

(a) s = ut + 0.5 a t^2

u = 0, t = 1.20, s = 2.10 giving a = 2.92 m/s/s

(b) F net = ma, m = 2.10, so F net = 2.10 x 2.92 = 6.132 N

Force down the slope = mg sin 30 = 10.3005 N

So force due to kinetic friction (up the slope) = 10.3005 - 6.132 = 4.1685 N

This force = coefficient x normal force = coefficient x mg cos (30), giving coefficient = 0.2336

(c) 4.1685 N, up the slope

(d) v = u + at = 0 + 2.92 x 1.20 = 3.5 m/s.