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How far does a block go up a frictionless ramp?

  1. Oct 18, 2004 #1
    A block is given an initial velocity of 3.00 m/s up a frictionless 22.0° incline. How far up the incline does the block slide before coming (momentarily) to rest? So basically, how far does it go up before the block starts falling.

    At first glance, this seemed really simple to me. Here's what I tried:

    The initial velocity in the Y direction would be 3sin(22) = 1.12 m/s

    Then I used the formula
    Vf2 = Vi2 + 2*(-9.8)*(delta Y)
    -->0 = 1.26 - 19.6(delta Y)
    so delta Y = .064 m.

    But that is straight up, and I need it along the incline. So, I did

    sin(22) = (.064)/x
    and x gives me .17 as the distance up the incline.

    However, neither .064m or .17m is correct. Can anyone tell me what I'm doing wrong? Am I totally off, or just missing a decimal or something?
  2. jcsd
  3. Oct 18, 2004 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Use that same kinematic equation ([itex]v_f^2 = v_i^2 + 2ax[/itex]) but use the components along the incline. Hint: Figure out the component of the acceleration due to gravity along the incline.
  4. Oct 19, 2004 #3
    surely you can do it by energy considerations - you know the KE at the start, and no GPE, then at the top there will be no KE momentarily, so you can work out h?
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