Question

Problema de coeficiente de rozamiento

Answer 1

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