# A trolleybus with a mass of 12 tons starts to move and within 5 seconds reaches a speed of 20 km / h.

A trolleybus with a mass of 12 tons starts to move and within 5 seconds reaches a speed of 20 km / h. What traction force does the trolleybus motor develop during this movement, if the movement is considered uniformly accelerated, and the friction force is taken equal to 2500N?

To find the value of the traction force that the motor of a given trolleybus should have developed, we will use the equality: m * a = Ftyag – Ftr, whence Ftyag = m * a + Ftr = m * V / t + Ftr.

The values of the variables: m is the mass of the trolleybus (m = 12 t = 12 * 103 kg); V is the achieved speed (V = 20 km / h = 5.56 m / s); t – the duration of the trolleybus acceleration (t = 5 s); Ftr – friction force (Ftr = 2500 N).

Calculation: Ftag = m * V / t + Ftr = 12 * 10 ^ 3 * 5.56 / 5 + 2500 ≈ 15.8 * 10 ^ 3 H = 15.8 kN.

Answer: The motor of this trolleybus was supposed to develop a traction force of 15.8 kN.

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