# A body weighing 50 kg under the action of a force of 200 N moves uniformly on a flat road.

July 25, 2021 | education

| **A body weighing 50 kg under the action of a force of 200 N moves uniformly on a flat road. what is the coefficient of friction?**

F = 200H

m = 50 kg

g = 10 m / s2

k-?

Since the body moves uniformly rectilinearly, then according to 1 Newton’s Law, the action on this body of all forces is compensated. The force with which the body is pulled F is compensated by the friction force Ftr, F = Ftr, Ftr = k * N, where k is the coefficient of friction, N is the support reaction. N = m * g, then Ftr = k * N = k * m * g. Since F = Ftr, then F = k * m * g. Let us express the coefficient of friction k = F / m * g. k = 200N / 50kg * 10m / s2 = 0.4.

Answer: coefficient of friction k = 0.4.

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