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.
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.
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