# An 8000 kg bus was traveling along a horizontal highway at a speed of 72 km / h.

**An 8000 kg bus was traveling along a horizontal highway at a speed of 72 km / h. After stopping the engine, the braking distance of the bus was 200 m. Determine the friction force acting on the bus during braking.**

m = 8000 kg.

V0 = 72 km / h = 20 m / s.

V = 0 m / s.

g = 10 m / s2.

S = 200 m.

Ftr -?

When braking a bus on a horizontal road, only 3 forces act on it:

m * g is the force of gravity, N is the reaction force of the road surface, Ftr is the friction force of the bus wheels on the road surface.

Let’s write 2 Newton’s law in vector form: m * a = m * g + N + Ftr.

Let’s write the expression of this law on the coordinate axes:

ОХ: m * a = Ftr.

OU: 0 = – m * g + N.

Ftr = m * a.

N = m * g.

We will assume that the braking of the bus was uniformly accelerated, so its acceleration is expressed by the formula: a = (V0 ^ 2 – V ^ 2) / 2 * S = V0 ^ 2/2 * S.

The formula for determining the friction force Ffr will take the form: Ffr = m * V0 ^ 2/2 * S.

Ftr = 8000 kg * (20 m / s) ^ 2/2 * 200 m = 8000 N.

Answer: when braking a bus, a friction force Ftr = 8000 N. acts on it.