What work does an electric locomotive do in 10 minutes, moving a train with a mass of 3000 tons along a horizontal path at a constant speed of 72 km / h, if the friction coefficient is 0.005?
t = 10 min = 600 s.
m = 3000 t = 3000000 kg.
g = 10 m / s2.
V = 72 km / h = 20 m / s.
μ = 0.005.
А – ?
Since the movement of the electric locomotive S coincides with the direction of its traction force Ft, its mechanical work A will be determined by the formula: A = Ft * S.
Since the electric train moves uniformly, the traction force of the train Ft is equal to the friction force Ftr: Ft = Ftr.
The friction force Ffr is expressed by the formula: Ffr = μ * m * g.
We express the displacement S by the formula: S = V * t.
A = μ * m * g * V * t.
A = 0.005 * 300000 kg * 10 m / s2 * 20 m / s * 600 s = 180,000,000 J = 180 MJ.
Answer: when moving the train, the electric train performs work A = 180 MJ.
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