Answer:
70 N
21°
1.1 m/s²
Step-by-step explanation:
Draw a free body diagram of the block. There are three forces:
Weight pulling straight down
Normal force pushing perpendicular to the incline
Friction force pushing parallel to the incline
Part 1
Sum the forces in the perpendicular direction:
∑F = ma
N − mg cos θ = 0
N = mg cos θ
The block is at rest, so F = N μs:
F = N μs
F = mg μs cos θ
F = (20 kg) (9.8 m/s²) (0.38) (cos 19°)
F = 70 N
Part 2
Sum the forces in the parallel direction (down the incline is positive):
∑F = ma
mg sin θ − F = 0
mg sin θ = N μs
mg sin θ = mg μs cos θ
tan θ = μs
θ = atan μs
θ = atan 0.38
θ = 21°
Part 3
Sum the forces in the parallel direction (this time, acceleration is not 0).
∑F = ma
mg sin θ − F = ma
mg sin θ − N μk = ma
mg sin θ − mg μk cos θ = ma
a = g (sin θ − μk cos θ)
a = (9.8 m/s²) (sin 24° − 0.32 cos 24°)
a = 1.1 m/s²