A 2.5kg block is on 53° inclined plane for which coefficient of kinetic friction is 0.25 and coefficient of static friction is 0.5. Calculate its acceleration given that:

Question

(a)it is intially at rest
(b)it is moving up the slope

Answer 1

M*g = 2.5 * 9.8 = 24.5 N. = Wt. of block.

Fp = 24.5*sin53 = 19.6 N. = Force parallel to the incline.

Fn = 24.5*Cos53 = 14.7 N. = Normal force.

Fs = us*Fn = 0.5 * 14.7 = 7.37 N.

Fk = uk*Fn = 0.25 * 14.7 = 3.69 N.

Answer 2

a. a = 0.

b. Fp-Fk = M*a.
a = (Fp-Fk)/M =