A 2.3 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 0.65 m/s. The incline is 1.6 m long.

Question

A 2.3 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 0.65 m/s. The incline is 1.6 m long.
What is the coefficient of friction?

Henry · Answer

Wb = m*g = 2.3kg * 9.8N/kg = 22.54 N. =
Wt. of block.

Fb = 22.54N @ 25o. = Force of block.
Fp = 22.54*sin25 = 9.53 N. = Force parallel to inclined plane.
Fv = 22.54*cos25 = 20.43 N. = Force perpendicular to plane = Normal.

V^2 = Vo^2 + 2ad.
a = (V^2-Vo^2)/2d
a = (0.4225-0)/3.2 = 0.132 m/s^2.

Fn = Fp-u*Fv = ma.
9.53-u*20.43 = 2.3*0.132 = o.304
-20.43u = 0.304-9.53 = -10.02 = -9.23
u = 0.45.