# When braking hard, the car reduces its speed from 50 m / s to 5 m / s in 15 seconds. What is the coefficient of sliding friction?

V0 = 50 m / s.
V = 5 m / s.
t = 15 s.
g = 10 m / s2.
μ -?
When braking a car, only 3 forces act on it: m * g is the force of gravity directed vertically downward, N is the reaction force of the road surface directed vertically upward, Ffr is the friction force directed in the opposite direction of the movement of the car.
Let’s write 2 Newton’s law in vector form: m * a = m * g + N + Ftr.
ОХ: m * a = Ftr.
OU: 0 = – m * g + N.
Ftr = m * a.
N = m * g.
The friction force Ffr is expressed by the formula: Ffr = μ * N = μ * m * g.
The acceleration of the car when braking a is found by the formula: a = (V0 – V) / t.
μ * m * g = m * (V0 – V) / t.
μ = (50 m / s – 5 m / s) / 10 m / s2 * 15 s = 0.3.
Answer: the coefficient of sliding friction is μ = 0.3.

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