Question

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?

Answer 1

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

Answer 2

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

• We'll begin by calculating the the normal reaction (N). This can be obtained as follow:

Mass (m) = 20 Kg

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

Normal reaction (N) =?

N = mg

N = 20 × 9.8

N = 196 N

• Finally, we shall determine the force of kinetic friction. This can be obtained as follow:

Coefficient of kinetic friction (µ) = 0.2

Normal reaction (N) = 196

Force of kinetic friction (F) =?

F = µN

F = 0.2 × 196

F = 39.2 N

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

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