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A 1.4-kg block slides freely across a rough surface such that the block slows down with an acceleration of â1.25 m/s2. what is the coefficient of kinetic friction between the block and the surface?

Respuesta :

Mass of the block = 1.4 kg

Weight of the block = mg = 1.4 × 9.8 = 13.72 N

Normal force from the surface (N) = 13.72 N

Acceleration = 1.25 m/s^2

Let the coefficient of kinetic friction be μ

Friction force = μN

F(net) = ma

μmg = ma

μg = a

μ = [tex]\frac{a}{g}[/tex]

μ = [tex]\frac{1.25}{9.8}[/tex]

μ = 0.1275

Hence, the coefficient of kinetic friction is: μ = 0.1275

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