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Box off a ramp

  1. Feb 23, 2015 #1
    1. The problem statement, all variables and given/known data

    http://imgur.com/VwoDF58
    2. Relevant equations
    How would I calculate the velocity the box gets off the ramp?

    I believe I need to use Vf = Vo + 2 a x

    a is gravity so it is -9.8, correct?

    but what do i put in as x? 5, which is the length of the ramp, or 4, the base, or 3 the height?
     
    Last edited: Feb 23, 2015
  2. jcsd
  3. Feb 23, 2015 #2

    Nathanael

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    The ramp is fixed and frictionless, right?

    Is gravity the only force acting on the box? In which direction is it accelerating?

    What is the distance the box travels?
     
  4. Feb 23, 2015 #3

    I'm not sure what you mean by fixed, but it is friction less.

    the box traveled 5 meters.
     
  5. Feb 23, 2015 #4

    Nathanael

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    I mean it can't move.

    What direction, what distance, and with what acceleration does the block move?
     
  6. Feb 23, 2015 #5
    the ramp is fixed.
    the box goes up the ramp.

    the distance it travels before flying off the ramp is 5 meters.

    and I have no idea how to calculate the acceleration. I assume it slows down near the top since it is going against gravity.
     
  7. Feb 23, 2015 #6

    Nathanael

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    It slows down the whole way. But at what rate?

    Have you tried drawing a free body diagram? What are all the forces acting on the block?
     
  8. Feb 23, 2015 #7
    Since it's frictionless, the only forces are gravity and normal force. correct? also the force from the initial push
     
  9. Feb 23, 2015 #8

    Nathanael

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    Right, but the force from the initial push is irrelevant. We can "start our clocks" immediately after that force stops; when the block is moving at 12 m/s

    What is the size and direction of the normal force? And what is the net force that results from adding gravity and the normal force together?
     
  10. Feb 23, 2015 #9
    direction of normal force is perpendicular to the surface.
    the size of the normal force is.... http://imgur.com/UOZntLH , the red line, correct?
     
  11. Feb 23, 2015 #10

    Nathanael

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    Yes but it's time to get into equations: What is the size of the normal force? More importantly, what is the size of the net force?
     
  12. Feb 23, 2015 #11
    I edited out my posit made no sense, to have Fg, we need the mass of the object, and that isn't given.
     
  13. Feb 23, 2015 #12
    but it should be cos theta = Fg⊥ to surface divided by Fg. correct?
     
  14. Feb 23, 2015 #13

    Nathanael

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    Maybe all hope is not lost, just continue as normal. The mass is hereby an arbitrary "m"

    Right, so that gives you |F| = |Fg|cosθ

    So what is the sum of the two vectors F+Fg ?
     
  15. Feb 23, 2015 #14
    |F⊥| = |Fg|cosθ= 7.84

    Fg = 9.8

    so adding them is 9.8 + 7.84 = 17.64

    is that correct?
     
  16. Feb 23, 2015 #15

    Nathanael

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    No, these are vectors. They have directions. You must add them component by component. Consider the component of gravity perpendicular to the ramp... How does it combine with the normal force? What net force do they produce in the direction perpendicular to the ramp? Then what is left of gravity?
     
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