A 25 kg box rests on an inclined plane which is 30o to the horizontal. The coefficient of sliding friction between the box and the plane is 0.30. Determine the acceleration of the box down the plane. (9.8 m/s2)

Answer 1

M*g = 25 * 9.8 = 245 N. = Wt. of the box

Fp = 245*sin30 = 122.5 N. = Force parallel to the incline.

Fn = 245*Cos30 = 212.2 N. = Force perpendicular to the incline.

Fk = u*Fn = 0.30 * 212.2 N. = 63.65 N. = Force of kinetic friction.

a = (Fp-Fk)/M*sin30 = (122.5-63.65)/12.5
= 4.71 m/s^2

Answer 2

Correction:

a = (Fp-Fk)/M = (122.6-63.65)/25 = 2.36
m/s^2