posted by Anonymous .
a 25.0kg box is released from rest on a rough inclined plane tilted at an angle of 33.5 degrees to the horizontal. the coefficient of kinetic friction between the box and the inclined plane is 0.200. determine the force of kinetic friction acting on the box. determine the accelleration of the box as it slides down the inclined plane
Wb = mg = 25kg * 9.8N/kg = 245 N. = Wt.
Fb=245N. @ 33.5 Deg. = Force of block.
Fp = 245*sin33.5 = 135.2 N. = Force in
parallel to inclined plane.
Fv = 245*cos33.5 = 204.3 N. = Force perpendicular to inclined plane.
a. Fk = u*Fv = 0.2 * 204.3 = 40.9 N =
Force of kinetic friction.
b. Fn = Fp - Fk = ma.
135.2 - 40.9 = 25a.
94.34 = 25a.
a = 94.34 / 25 = 3.77 m/s^2.