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.
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.
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