(1) When the train works at rated power, when traction equals resistance, that is, when the acceleration of the train is zero and the speed reaches the maximum value vm, then:

vm=PF=PFf=Pkmg= 12？ Postgraduate students.

(2) When V < VM, the train accelerates, when v=v2= 10? M/s，F2=Pv2=6× 104？ ordinary

According to Newton's second law: a2=F2? Ffm=0.02？ m/s2。

(3) When v=36? km/h= 10？ M/s, the car is driving at a constant speed, and the actual engine power is P=Ffv=5× 105? W.

(4) According to Newton's second law, the traction force F ′ = FF+MA = 3×105? N, in this process, the speed increases and the engine power increases.

At rated power, the rotating speed is VM ′, that is, VM ′ = PF ′ = 2? M/s

According to VM ′ = at, t = VM'a = 4? south

According to kinetic energy theorem:

W-Ffx= 12m(at)2

X= 12at2。

Have to? w = ff× 12at 2+ 12m(at)2 = 1.2× 106j

A:

(1) The maximum speed of the train running on the horizontal track is12m/s;

(2) On the horizontal track, when the engine works at rated power P and the running speed v= 10m/s, the instantaneous acceleration A of the train is 0.02m/S2；; ;

(3) The actual engine power P' is 5×105 w; When driving at a constant speed of 36 km/h on a horizontal track;

(4) If the train starts from a stationary state and accelerates at a constant speed with an acceleration of 0.5m/s2, the longest time of this process is 4s, and the work done by the engine during this period is1.2xx106j.