The friction force acting on a sled with a mass of 80 kg sliding on a horizontal road is 16 N.

| July 13, 2021 | education

The friction force acting on a sled with a mass of 80 kg sliding on a horizontal road is 16 N. The coefficient of sliding friction of steel on ice is.

Ftr = 16 N.

m = 80 kg.

g = 9.8 m / s2.

μ -?

The sliding friction force Ftr is determined by the formula: Ftr = μ * N, where μ is the sliding friction coefficient, N is the surface reaction force, the force with which the surface presses on the sled.

According to Newton’s 3 law, the force N with which the surface acts on the sled is equal to the force P with which the sled presses on the surface, that is, the weight of the sled: N = P.

We will express the weight of the sleigh P by the formula: P = m * g, where m is the mass of the sleigh, g is the acceleration of gravity.

Ftr = μ * m * g.

μ = Ftr / m * g.

μ = 16 N / 80 kg * 9.8 m / s2 = 0.02.

Answer: the coefficient of sliding friction is μ = 0.02.



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