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Velocity at bottom of ramp

  1. Sep 17, 2011 #1
    1. The problem statement, all variables and given/known data

    Starting from rest an 11kg block slides 11.9m down a frictionless ramp (inclined at 30° from the floor) to the bottom. The block then slides an additional 26.3m along the floor before coming to a stop. The acceleration of gravity is 9.8 m/s^2. Find the speed of the block at the bottom of the ramp.


    2. Relevant equations

    Vf^2=vi^2+2a(d-do)

    Conservation of energy: Ebefore=Eafter

    KE=.5mv^2

    PE=mgh


    3. The attempt at a solution

    I tried solving this two differant ways, and both gave me the same answer, which was wrong. First I tried kinematics.

    Vf^2=vi^2+2a(d-do)
    Vf=sqrt(2*9.8/sin30*11.9)

    (I used 9.8/sin30 because 9.8 is the vertical component, so 9.8/sin30 is the acceleration down the ramp)

    Vf=21.598


    The other way I tried used conservation of energy.

    h=11.9/sin30=23.8

    PE=KE
    mgh=.5mv^2
    gh=.5v^2
    v=sqrt(9.8*23.8/.5)
    v=21.598

    I don't see where I went wrong in my logic, so I would greatly appreciate some guidance. Thank you!
     
    Last edited: Sep 17, 2011
  2. jcsd
  3. Sep 17, 2011 #2

    PeterO

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    In the first case you have used 9.8/sin30. It should be 9.8*sin30
    In the second case you used h/sin30. That should have been h*sin30.
     
  4. Sep 17, 2011 #3
     
  5. Sep 18, 2011 #4

    PeterO

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    Homework Helper

     
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