# A car with a mass of 1000 kg, moving at a speed of 36 km / h, starts to brake. Determine the friction force

**A car with a mass of 1000 kg, moving at a speed of 36 km / h, starts to brake. Determine the friction force and the distance traveled by the car to a stop, if the coefficient of friction is 0.5.**

m = 1000 kg.

V0 = 36 km / h = 10 m / s.

V = 0 m / s.

g = 10 m / s2.

μ = 0.5.

Ftr -?

S -?

When braking in the horizontal direction, only the friction force Ffr acts on the car. 2 Newton’s law in vector form will have the form: m * a = Ftr + m * g + N, where m * g is the force of gravity, N is the reaction force of the road surface.

ОХ: m * a = Ftr.

OU: 0 = N – m * g.

Ftr = m * a.

N = m * g.

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

Ftr = 0.5 * 1000 kg * 10 m / s2 = 5000 N.

a = μ * g.

The braking path S is expressed by the formula: S = (V0 ^ 2 – V ^ 2) / 2 * a = (V0 ^ 2 – V ^ 2) / 2 * μ * g.

S = ((10 m / s) ^ 2 – (0 m / s) ^ 2) / 2 * 0.5 * 10 m / s2 = 10 m.

Answer: the frictional force during braking is Ftr = 5000 N, the braking distance

S = 10 m.