A box of mass 52 kg starts from rest to slide down a ramp that makes an angle of 16.6 degrees with respect to the horizontal. 4.1 seconds later, it has covered a distance of 3.98 meters. What is the coefficient of kinetic friction?

Answer 1

Fp=mg*sin16.6 = 509.6*sin16.6=145.6 N. =
Force parallel to the ramp.

Fn = 509.6*cos16.6 = 488.4 N. = Normal force = Force perpendicular to the ramp.

d = Vo*t + 0.5a*t^2 = 3.98 m.
0 + 0.5a*4.1^2 = 3.98
8.405a = 3.98
a = 0.474 m/s^2.

Fp-Fk = m*a
145.6 - Fk = 52*0.474
-Fk = 24.62-145.6 = -121.0
Fk = 121.0 N.=Force of kinetic friction.

u = Fk/Fn = 121/488.4 = 0.248 = Coefficient of kinetic friction.