A 3.00-kg block starts from rest at the top of a 25.5° incline and slides 2.00 m down the incline in 1.75 s.

Question

(a) Find the acceleration of the block.
m/s2

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


(c) Find the frictional force acting on the block.
N

(d) Find the speed of the block after it has slid 2.00 m.
m/s

Answer 1

a. V = d/t = 2.0m/1.75s. = 1.143 m/s.

V^2 = Vo^2 + 2a*d
a = (V^2-Vo^2)/2d
a = (1.143^2-o)/4 = 0.327 m/s^2

b. Wb = m*g = 3kg * 9.8N/kg = 29.4 N. =
Weight of block.

Fp = 29.4*sin25.5 = 12.66 N. = Force
parallel to incline.

Fv = 29.4*cos25.5 = 26.54 N. = Force
perpendicular to incline.

Fp-Fk = m*a
12.66-u*26.54 = 3 * 0.327
-u*26.54 = 0.981-12.66 = -11.68
uk = 0.440

c. Fk = uk * Fv = 0.440 * 26.54=11.68 N.

d. V = 2m/1.75s = 1.143 m/s.