An engine is used to pull a train of two cars out of a mine. The floor slopes upward at an angle of 24o. Each car has a mass of 0.8x104 Kg and normally travels without friction on the tracks. The engine can exert a maximum force of 1.4x105 N on car A. If the engineer again throttles back so that the force exerted by the engine on car A decreases at the constant rate of 3.9 N per second, how long before the train stoprs moving up the track? Assume the original speed was 2.9 m/s

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  1. Force from engine = 1.4*10^5 - 3.9 t

    The cars do not move apart, so can be considered one rigid body twice as massive. m = 1.6 * 10^4 kg

    Slope = 24 deg so component of weight down track = m g sin 24 = 1.6*10^4 * 9.8 * .407
    = 6.38 *10^4 N down track

    F = m a
    up track
    F = 1.4*10^5 - 3.9 t -6.38*10^4 = (1.6*10^4) a
    1.6 * 10^4 a = 7.62 * 10^4 -3.9 t
    a = d^2x/dt^2 = 4.76 - 2.44*10^-4 t

    now if it were constant acceleration, I would assume that v = Vo + a t
    so I will try something like that to solve this differential equation

    v = Vo + B t + C t^2
    then a = dv/dt = B + 2C t
    well that is kind of encouraging
    I will use B = 4.76 and 2C = -2.44 *10^-4
    v = 2.9 + 4.76 t - 1.22 *10^-4 t^2
    the train stops when v = 0
    0 = 2.9 + 4.76 t - 1.22 *10^-4 t^2
    solve that quadratic equation for t

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  2. Check my arithmetic (and your given decrease of 3.9 N/second) because I get about 3.9*10^4 seconds which is around 10 hours.

    I think you have a typo, throttle should be backed off like 3.9*10^4 N/s or something. It takes forever to back off from 1.4*10^5 N at only 3.9 N/s

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