A car weighing 1.5 tons begins to move with an acceleration of 0.5 m / s ^ 2. During the movement, a resistance force

A car weighing 1.5 tons begins to move with an acceleration of 0.5 m / s ^ 2. During the movement, a resistance force of 500 n acts on the car. Determine the traction force if the driver forgot to remove the handbrake, which has a friction force of 100N?

m = 1.5 t = 1500 kg.

a = 0.5 m / s2.

Fcopr = 500 N.

Ftr = 100 N.

Fт -?

Let’s write Newton’s 2 law for a car: m * a = Fр, where m is the mass of the car, a is the acceleration of the car, Fр is the resultant of all forces that act on the car.

We will consider the movement of the car in the horizontal direction.

The resultant of all forces Fr will be determined by the vector sum: Fr = Ft + Fcopr + Ftr, where Ft is the engine thrust force, Fcopr is the drag force, Ftr is the friction force.

Since the friction force Ftr and the resistance Fcopr are directed in the opposite direction of the body’s movement, then: Fr = Ft – Fcopr – Ftr.

m * a = Fт – Fcopr – Ftr.

Ft = m * a + Fcopr + Ftr.

Ft = 1500 kg * 0.5 m / s2 + 500 N + 100 N = 1350 N = 1.35 kN.

Answer: the traction force of the car is Ft = 1.35 kN.