Question

A block with a mass of 5.0 kg slides down a 370 incline as 1 point
shown below with an acceleration of 5.6 m/s2 (sin 370 =
0.6, cos 370 = 0.8). The coefficient of kinetic friction
between the block and the inclined surface is 0.050. The
normal force, FN applied by the inclined plane on the block
is nearly: *
50N
40 N
30 N
20N

Answer 1

Answer:

40 N

Explanation:

Draw a free body diagram of the block.  The block has three forces acting on it: normal force pushing perpendicular to the incline, weight force pulling down, and friction pushing parallel to the incline.

Sum of the forces perpendicular to the incline:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

Given m = 5.0 kg and cos θ = 0.8:

N = (5.0 kg) (9.8 m/s²) (0.8)

N ≈ 40 N