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Block on a frictionless incline

  1. Oct 14, 2007 #1
    1. The problem statement, all variables and given/known data

    A block of mass m = 2.00 kg is released from rest at h = 0.500 m from the surface of a table, at the top of a θ = 35.0° incline as shown below. The frictionless incline is fixed on a table of height H = 2.00 m.

    How much time has elapsed between when the block is released and when it hits the floor?

    2. Relevant equations

    3. The attempt at a solution

    I was required to find the acceleration of the block as it moves down the incline. I found that to be 5.621 m/s^2

    I then found the velocity at the point where the block leaves the incline to be 3.13 m/s.

    Then I found the distance that the block travels from when it leaves the table to when it hits the ground. This I found to be 1.234 m.

    Then I attempted to find the total time of the blocks movement:

    Velocity in the y direction:
    Vyo = 3.13 Cos(55)
    Vyo = -1.7953 m/s

    Time for the block from the edge of the table to the floor:
    Y = Yo + Vyo - 1/2at^2
    0 = 2m - 1.7953t - 1/2(9.8)t^2
    t = .48143 sec

    Velocity in the x direction:
    Vxo = 3.13 Sin(55)
    Vxo = 2.56395 m/s

    Vf = Vo + at

    2.56395 = 0 + 5.621(t)
    t = .456 sec

    So, the total time would be the two times added together:
    .48143 sec + .456 sec = .93743 sec

    I submitted this answer and it said it was within 10% of the correct answer....Am I making some sort of rounding error or is there some mistake in my calculations? Thanks for any help!

    Attached Files:

  2. jcsd
  3. Oct 14, 2007 #2


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    Homework Helper

    "2.56395 = 0 + 5.621(t)
    t = .456 sec"

    why are you using 2.56395 instead of 3.13 here?
  4. Oct 14, 2007 #3
    That would explain my mistake. I was using the horizontal component of the velocity. When I use 3.13 I get 1.03827 sec, which is correct. Thanks a lot learningphysics!
  5. Oct 14, 2007 #4


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    Homework Helper

    no prob.
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