# 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.