# A car with a mass of 5.5 tons, moving at a speed of 30 km / h, is stopped by brakes on a path of 5 m.

**A car with a mass of 5.5 tons, moving at a speed of 30 km / h, is stopped by brakes on a path of 5 m. Determine the braking force, considering the movement in the braking period to be equally slow. The coefficient of friction of the wheels is 0.18.**

m = 5.5 t = 5500 kg.

g = 10 m / s2.

V0 = 30 km / h = 8.3 m / s.

V = 0 m / s.

S = 5 m.

μ = 0.18.

Ftr -?

When braking the car 2, Newton’s law in vector form will have the form: m * a = m * g + N + Ftr + Ft, where m * g is the force of gravity, N is the reaction force of the road to the wheels, Ftr is the friction force of the wheels on the road , Fт – braking force.

ОХ: m * a = Ftr + Ft.

OU: 0 = – m * g + N.

Ft = m * a – Ftr.

N = m * g.

The friction force Ffr is expressed by the formula: Ffr = μ * N = μ * m * g.

Ftr = 0.18 * 5500 kg * 10 m / s2 = 9900 N.

The acceleration of the car during braking and we express the formula: a = (V0 ^ 2 – V ^ 2) / 2 * S = V02 / 2 * S.

Ft = m * V0 ^ 2/2 * S – Ftr.

Ft = 5500 kg * (8.3 m / s) ^ 2/2 * 5 m – 9900 N = 27989.5 N.

Answer: the braking force is Ft = 27989.5 N.