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The truck, the box and the slide

  1. Sep 25, 2009 #1
    problem is attached as image...

    i asked on yahoo answers, received a reply but is wrong...

    it is very challenging (in my opinion)..
    I spent over an hour without reaching even a wrong option..
     

    Attached Files:

  2. jcsd
  3. Sep 25, 2009 #2

    Doc Al

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    Staff: Mentor

    What do you think? Show what you've done so far.

    Hints: What's the maximum acceleration of the box? Figure out how far the box moves with respect to the ground.
     
  4. Sep 25, 2009 #3
    http://answers.yahoo.com/question/i...nkoSvGAazKIX;_ylv=3?qid=20090925002613AAbpiRt

    Dr D's answer is correct:

    The block will slide once the deceleration of the truck exceeds μg.
    Once the block begins to slide, its absolute deceleration is constant at μg.
    So the stopping distance of the block is v^2 / (2μg) = 40.77 m.

    And the block is allowed to stop within 3 m more than the truck. So the minimum stopping distance of the truck is 37.77 m.

    If you use g = 9.81, you get 37.77 m
    If g = 10, then you get 37 m.

    but his answer is a bit short, and hard to understand...
     
  5. Sep 26, 2009 #4

    Doc Al

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    Staff: Mentor

    Rather than hunting for answers, your time might be better spent trying to figure it out on your own. Use the hints I gave and your knowledge of Newton's laws.
     
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