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第一种类型 [例] 总质量为M的列车,沿水平直轨道匀速前进.其质量为m的末节车厢突然脱钩,从脱钩到司机发觉机车已行驶的路程为l.司机一发觉就切断动力,除去牵引力.设运动的阻力与车重成正比,除去牵引力前机车的牵引力是恒定的.试求列车的前后两部分最终相距s=? 解:用功能补偿法解.假设在脱钩处机车同时撤除牵引力,则由于这两部分的初速度相同,做匀减速运动的加速度也相同(若阻力为车重的k倍,就可求得两部分的加速度都为-kg),它们从脱钧到停止通过的路程应相同.但现在列车的前部比末节车厢多行一段路程s才停下,就要多克服阻力做功k(M-m)gs,这应由机车牵引力F_对它多做的功W_=F_l来补偿.注意到F_=kMg,可得
The first type [example] train with a total mass of M travels along a horizontal straight track at a uniform speed. The mass car with a mass of m suddenly decouples from the hook. From decoupling to the driver, the distance traveled by the locomotive is found to be l. The driver cuts off the power as soon as he detects it. Removal of traction. The resistance of the movement is proportional to the weight of the vehicle. The traction of the locomotive before removal of the traction is constant. Try to find the final distance between the two parts of the train. s=? Solution: Use the functional compensation method to solve. Suppose the locomotive is at the decoupling position. Remove the traction force, because the initial velocity of the two parts is the same, the acceleration of the uniform deceleration movement is also the same (if the resistance is k times the weight of the vehicle, it can be obtained that the acceleration of both parts is -kg), and they are from dislocation to The stopping distance should be the same. However, now the front of the train will stop more than the last one for a while, it is necessary to overcome the resistance to do work k(Mm)gs, which should be done by the locomotive traction force F_. W_=F_l to compensate. Note that F_=kMg, available