question_answer

  A 8000 kg engine pulls a train of 5 wagons, each of 2000 'kg, along ahorizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate (a) the net acceleration force, (b) the acceleration of the train and (c) the force of wagon 1 on wagon 2. Or The driver of train A travelling at a speed ofapplies brakes and retards the train uniformly. The train stops in 5 s. Another train B is travelling on the parallel with a speed of Its driver applies the brakes and the train retards uniformly; train B stops in 10 s. Plot speed-time graphs for both the trains on the same axis. Which of the trains travelled farther after the brakes were applied?

Answer:

  (a) Net Accelerating Force = Force Exerted by Engine - Friction Force      (Here frictional force is subtracted because it opposes the motion) = 40000 - 5000 = 35000N = 3.5104 N                                                                                                                                                                                       (1) (b) From Newton's second law of motion, Accelerating Force = Mass of the train x Acceleration of train Mass of Train = Mass of Engine + Mass of all Wagons = 8000 + 52000 = 8000 + 10000 = 18000 kg                                                                                                                                                                                           (1)  Acceleration                                                                                                                                                                                (1) (c) Force of wagon 1 on wagon 2 = Mass of the Engine x Acceleration                                                                                                                                        (1) = 15600 N                                                                                                                                                                                            (1) Or For train A, the initial velocity,                                                                                                                                   (1/2) Final velocity, v = 0 and time, t = 5 For train                                                                                                           (1/2) Speed-time graph for train A and B are shown in figure.                                                                                                                         Distance travelled by train A = Area under straight line RS                                                                                                                                                                         (1) Distance travelled by train B = Area under                                                                                                                                                                   (1) Therefore train B travelled more distance than train A.                                                                                  (1/2)