A body weighing 10 kg is pulled horizontally and evenly on a horizontal surface under the action of a force

A body weighing 10 kg is pulled horizontally and evenly on a horizontal surface under the action of a force of 20 N directed horizontally. What is the value of the coefficient of friction?

A body weighing 10 kg is pulled horizontally and evenly on a horizontal surface under the action of a force of 20 N directed horizontally. What is the value of the coefficient of friction?

m = 10 kg.

g = 10 m / s ^ 2.

F = 20 N.

V = const.

μ -?

According to 1 Newton’s law, a body is in a state of uniform rectilinear motion or a state of rest, if it is not acted upon by forces or the action of forces is compensated.

The force of gravity m * g is compensated by the reaction force of the support N, the friction force Ffr is compensated by the force F with which the body is pulled along the floor.

N = m * g.

F = Ftr.

The friction force Ffr is determined by the formula: Ffr = μ * N, where μ is the coefficient of friction.

Ftr = μ * m * g.

μ = Ftr / m * g.

μ = 20 N / 10 kg * 10 m / s ^ 2 = 0.2.

Answer: the coefficient of friction is μ = 0.2.