Find the braking force acting on a car weighing 3.2 tons if the speed at the beginning

Find the braking force acting on a car weighing 3.2 tons if the speed at the beginning of braking was 9 m / s and the braking distance was 67.5 m.

Data: m (vehicle weight) = 3.2 t; V0 (vehicle speed at the beginning of braking) = 9 m / s; S (braking distance) = 67.5 m.

SI system: m = 3.2 t = 3200 kg.

Option 1.

Vehicle acceleration:

S = (V ^ 2 – V0 ^ 2) / 2a = -V0 ^ 2 / 2a.

a = -V0 ^ 2 / 2S = -9 ^ 2 / (2 * 67.5) = -0.6 m / s ^ 2.

Effective braking force:

F = | m * a | = 3200 * 0.6 = 1920 N.

Option 2.

Braking force work:

A = F * S = m * V0 ^ 2/2.

F = m * V0 ^ 2 / 2S = 3200 * 9 ^ 2 / (2 * 67.5) = 1920 N.

Answer: The braking force is 1920 N.