A 4 kg block slides down a rough inclined plane inclined at 30° with the horizontal. Determine the coefficient of kinetic friction between the block and the surface if the block has an acceleration of 1.2 m/s2.

Answer 1

Wb = m*g = 4kg * 9.8N/kg = 39.2 N. = Wt.
0f block.

Fb = 39.2N. @ 30o = Force of block.
Fp=39.2*sin30 = 19.6 N.=Force parallel to plane.
Fv = 39.2*cos30 = 33.9 N. = Force perpendicular to plane.

Fk = u*Fv = u*33.9.

Fn = Fp-Fk = m*a.
19.6 - 33.9u = 4*1.2 = 4.8
-33.9u = 4.8 - 19.6 = -14.8
u = 0.437 = Coefficient of kinetic friction.

Answer 2

The block shown in Figure 6 has mass 3500 g and lies on a plane which tilted at an angle θ = 22° to the horizontal. The effective coefficient of kinetic friction is 0.20. Determine the acceleration of the block as it slides down the plane.