(a) Acceleration: 2.24 m/s²
(b) Coefficient of kinetic friction: 0.30
(c) Frictional force: 8.07 N
(d) Speed after 1.92 m: 2.88 m/s
How can you find the magnitude of the acceleration of the block?
mg sin(θ) - f k = ma
m = 3.12 kg (mass of the block)
g = 9.81 m/s² (acceleration due to gravity)
θ = 30° (angle of the incline)
f k = coefficient of kinetic friction × normal force (friction force)
a = acceleration of the block (unknown)
Normal force = mg cos(θ)
= 3.12 kg × 9.81 m/s² × cos(30°)
= 26.59 N
a = (mg sin(θ) - f k) / m
a = (26.59 N × sin(30°) - f k) / 3.12 kg
(b) Coefficient of kinetic friction:
d = 1/2 at²
1.92 m = 1/2 a × 1.70 s²
a = 2.24 m/s²
2.24 m/s² = (26.59 N × sin(30°) - f k) / 3.12 kg
f k = 26.59 N × sin(30°) - 3.12 kg × 2.24 m/s²
f k ≈ 8.07 N
μ k = f k / (mg cos(θ))
μ k ≈ 8.07 N / (3.12 kg × 9.81 m/s² × cos(30°))
μ k ≈ 0.30
(c) f k ≈ 8.07 N.
(d) Speed of the block after 1.92 m:
v² = u² + 2as
v² = 0² + 2 × 2.24 m/s² × 1.92 m
v² = 8.32 m²/s²
v = √8.32 m²/s² ≈ 2.88 m/s