An electric train weighing 10 ^ 6 kg begins to move uniformly and within 1 minute reaches a speed of 108 km / h.

An electric train weighing 10 ^ 6 kg begins to move uniformly and within 1 minute reaches a speed of 108 km / h. Determine the traction force of the electric train if the friction coefficient is 0.02.

m = 10 ^ 6 kg.

g = 10 m / s2.

t = 1 min = 60 s.

V0 = 0 m / s.

V = 108 km / h = 30 m / s.

μ = 0.02.

F -?

Let us write Newton’s 2 law in vector form for an electric train: m * a = F + Ftr + m * g + N, where m is the mass of the train, a is the acceleration of the train, F is the traction force of the train, Ftr is the friction force, m * g – gravity, N is the reaction force of the rails.

ОХ: m * a = F – Ftr.

OU: 0 = – m * g + N.

F = m * a + Ftr.

N = m * g.

Let us write down the formula for the acceleration of the body: a = (V – V0) / t.

Since V0 = 0 m / s, then a = V / t.

The friction force Ffr is determined by the formula: Ffr = μ * N = μ * m * g.

F = m * V / t + μ * m * g = m * (V / t + μ * g).

F = 10 ^ 6 kg * (30 m / s / 60 s + 0.02 * 10 m / s2) = 0.7 * 106 kg.

Answer: the traction force of the train is F = 0.7 * 106 kg.




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