The body begins to brake with a speed of 15 m / s, when braking, the body travels a certain distance in 5 s

The body begins to brake with a speed of 15 m / s, when braking, the body travels a certain distance in 5 s, the body weight is 700 kg, you need to find the coefficient of friction and the friction force if the acceleration of the body is 3 m / s.

V0 = 15 m / s.

t = 5 s.

m = 700 kg.

g = 9.8 m / s ^ 2.

a = 3 m / s ^ 2.

Ftr -?

μ -?

Let us write Newton’s 2 law in vector form: m * a = Ftr + m * g + N, where m is the mass of the body, a is the acceleration of the body, Ftr is the friction force, m * g is the force of gravity, N is the reaction force of the plane.

ОХ: m * a = Ftr.

OU: 0 = m * g – N.

Ftr = 700 kg * 3 m / s ^ 2 = 2100 N.

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 = 2100 N / 700 kg * 9.8 m / s ^ 2 = 0.3.

Answer: the friction force is Ffr = 2100 N, μ = 0.3.