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Homework Help: Will the train's cargo slam into the front of the boxcar?

  1. Apr 10, 2012 #1

    A train travelling at 45MPH comes to a stop in 31 feet, experiencing a uniform deceleration.
    Inside one of its flat-bed boxcar's is a steel block that sits 9 feet from the front wall of its container.


    If the load indeed strikes the wall of the car (which I've determined it does), please calculate its speed relative to the train as the moment of impact.

    • Coefficient of static friction = 0.3
    • Coefficient of kinetic friction = 0.25
    • Initial velocity of the train = 45MPH = 20.1168m/s
    • Stopping distance of the train = 31' = 9.4488m
    • Block's distance from the forward wall = 9' = 2.7432m

    I'm half asleep at the moment so I'm not exactly confident about my answers but so far I've determined that if the train had to stop within 31 feet, it's deceleration would be -21.414ms-2. Again, that's probably not even close.

    And I'm not really sure where to go from here...
    Last edited by a moderator: Apr 10, 2012
  2. jcsd
  3. Apr 10, 2012 #2


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    Your acceleration of the train looks good, in a direction opposite it's motion . Now you need to calculate the acceleration of the crate, with respect to the ground, using newton 2 after first identifying the net force acting on it. Noe find the relative acceleration, and the relative speed in the manner you used to find the trains acceleration . You don't have to convert to metric it's just another source for math errors.
  4. Apr 10, 2012 #3
    I'm still confused. I guess my problem is finding the relative acceleration.

    Is it simply equal to the sum of the acceleration of the train and the kinetic friction?
  5. Apr 10, 2012 #4
    I also don't understand when I'm supposed to use static friction or kinetic friction in this problem.
  6. Apr 10, 2012 #5


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    Well, no, that would sort of be like adding apples and oranges....acceleration (m/s^2 or ft/s^2) and force (N or lb) are not of the same units. Relative accelerations tend to get confusing....basically, if a crate is accelerating at 2 m/s^2 to the left, with respect to the ground, and the truck is accelerating at 10 m/s^2 to the left, with respect to the ground, then the acceleration of the crate with respect to the truck is 8m/s^2 to the right (V_ct = V_c - V_t, vector subtraction, watch plus and minus signs!)). But first you must find the acceleration of the crate in this problem, w.r.t. the ground, by identifying the lone force (and its direction) acting on it in the horizontal direction, and using Newton's 2nd Law. Note that since the crate is sliding with respect to the truck, the friction force must be kinetic, not static.
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