A 20kg block slides a cross a horizontal surface. If the coefficient of kinetic friction between the block and the surface is 0.20, what is the force of kinetic friction that acts on the block?

The force of kinetic friction acting on the block of mass 20 Kg having a coefficient of kinetic friction of 0.2 is 39.2 N

Mass (m) = 20 Kg

Acceleration due to gravity (g) = 9.8 m/s²

N = mg

N = 20 × 9.8

Coefficient of kinetic friction (µ) = 0.2

Normal reaction (N) = 196

F = µN

F = 0.2 × 196

Therefore, the force of kinetic friction acting on the block is 39.2 N

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Compute the block's weight:

w = m g = (20 kg) (9.8 m/s²) = 196 N

The net vertical force is 0, so the normal force has the same magnitude:

n + (-w) = 0   →   n = 196 N

The frictional force (mag. f ) is proportional to the normal force by a factor of µ = 0.20 such that

f = 0.20 (196 N) = 39.2 N ≈ 39 N

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