A wooden block is evenly pulled along a horizontal surface applying a force of 1H to determine the sliding friction coefficient if the weight of the block is 200 g
According to Newton’s second law:
ma = F – Ftr, m – body weight (m = 200 g = 0.2 kg), a – acceleration of the body (with uniform movement a = 0 m / s²), F – acting force (F = 1 N), Ftr – friction force (Ftr. = Μ * N, μ – friction coefficient, N – support reaction force (for a horizontal surface N = Fт (gravity) = m * g, g – free fall acceleration (g = 10 m / s²) )).
0 = F – Ftr.
F = μ * m * g.
μ = F / (m * g) = 1 / (0.2 * 10) = 0.5.
Answer: The coefficient of sliding friction is 0.5.
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