Question

A 2.6 kg block slides down a 25° inclined plane with constant acceleration. The block starts from rest at the top. At the bottom, its velocity is 0.75 m/s. The incline is 1.6 m long.
(a) What is the acceleration of the block?

(b) What is the coefficient of friction?

Answer 1

a. Wb = M*g = 2.6kg * 9.8N/kg=25.48 N. =
Weight of block.

Fp = 25.48*sin25 = 10.77 N. = Force
parallel to the inclined plane.
Fn = 25.48*cos25 = 23.09 N. = Normal Force = Force perpendicular to the
incline.

a=V^2-Vo^2)/2d
a = (0.75^2-0)/2.6 = 0.2163 m/s^2.

b. Fp-Fk = m*a
10.77-Fk = 2.6 * 0.2163 = 0.5625
Fk = 10.77 - 0.5625 = 10.21 N. = Force
of kinetic friction.

u = Fk/Fn = 10.21/23.09 = 0.4421

Answer 2

Correction:
a = (0.75^2-0)/3.2 = 0.1758 m/s^2.

b. Fp-Fk = m*a
10.77-Fk = 2.6*0.1758 = 0.4570
Fk = 10.77 - 0.4570 = 10.31 N

u = Fk/Fn = 10.31/23.09 = 0.4466