A car with a mass of 2 t moves on a horizontal surface with an acceleration of 2.5 m / s2.

A car with a mass of 2 t moves on a horizontal surface with an acceleration of 2.5 m / s2. Determine the traction force developed by the car engine if the coefficient of friction is 0.1.

m = 2 t = 2000 kg.

a = 2.5 m / s2.

g = 10 m / s2.

μ = 0.1.

Fт -?

According to Newton’s 2 law, the resultant of all forces Fр that act on a body is equal to the product of the mass of this body m by its acceleration a: m * a = Fр.

The resultant of all forces Fr is equal to the vector sum of all forces that act on the car: Fr = Ft + Ftr + m * g + N.

ОХ: m * a = Fт – Fтр.

DT: 0 = -m * g + N.

Ft = m * a + Ftr.

N = m * g.

We see that the traction force Ft is used to impart acceleration to the car and overcome the friction force.

Ftr = μ * N = μ * m * g.

Fт = m * a + μ * m * g = m * (a + μ * g).

Ft = 2000 kg * (2.5 m / s2 + 0.1 * 10 m / s2) = 7000 N.

Answer: the car has a traction force Ft = 7000 N.