If you take the positive horizontal direction to be to the right (the same direction as F₂), then by Newton's second law, the block has
• a net horizontal force of
∑ F = F₁ cos(-50°) + F₂ - f = m a
• a net vertical force of
∑ F = F₁ sin(-50°) + n - m g = 0
where
• f = µ n = magnitude of friction
• µ = coefficient of kinetic friction
• n = magnitude of the normal force
• m = 20 kg
• a = acceleration of the block
Solve for n :
n = m g + F₁ sin(50°)
n = (20 kg) (9.80 m/s²) + (50 N) sin(50°)
n ≈ 234.302 N
If the block accelerates uniformly from rest with acceleration a, then this acceleration is equal to its average, given by
a = ∆v / ∆t = (9 m/s - 0) / (3 s) = 3 m/s²
Solve for f :
f = F₁ cos(50°) + F₂ - m a
f = (50 N) cos(50°) + 80 N - (20 kg) (3 m/s²)
f ≈ 52.1394 N
Solve for µ :
µ = f / n
µ ≈ (52.1394 N) / (234.302 N)
µ ≈ 0.22253 ≈ 0.22
