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Using force equations and kinematic equations

  1. Sep 29, 2007 #1
    1. The problem statement, all variables and given/known data

    A 1.0kg wood block is launched up a wooden ramp that is inclined at a 38 degree angle. The block's initial speed is 14 m/s. (Use uk = .20 for the coefficient of kinetic friction for wood on wood.) a.) What vertical height does the block reach above its starting point? b.) What speed does it have when it slides back down to its starting point?

    2. Relevant equations

    F = ma
    Frictional force = uk(n)
    kinematic equations

    3. The attempt at a solution
    I broke the weight into its components. Then I found the forces for each direction. I then solved for the normal force, and got 59.249N. I then substituted that in for the equation in the x-direction to find the acceleration. I got -67.179 m/s^2. I then tried using Vxf^2 = Vxi^2 + 2(ax)(delta x).
     
  2. jcsd
  3. Sep 29, 2007 #2
    I cannot see friction in your equations(or method), and normal force cannot be 59 N. It is much smaller
     
  4. Sep 29, 2007 #3
    Normal - Wy = m(ay) or Normal - mgcos (theta) = m(ay) right?
     
  5. Sep 29, 2007 #4
    m(ay) is 0
    so
    N = mgcos theta

    N = 1 * g * cos (t)

    and how u got >10?
     
  6. Sep 29, 2007 #5
    The normal force is 7.73N. So now I substitute the normal force into -fricitional force-Wx = m(ax).
     
  7. Sep 29, 2007 #6

    learningphysics

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    Are you allowed to use work/energy for this problem?
     
  8. Sep 29, 2007 #7
    I don't think so. We haven't gone over work or energy yet.
     
  9. Sep 29, 2007 #8

    learningphysics

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    yup. this looks right.
     
  10. Sep 29, 2007 #9
    For the acceleration, I got -7.586 m/s^2. I tried using Vxf^2 = Vxi^2 + 2(ax)(delta x), but it didn't work.
     
  11. Sep 29, 2007 #10

    lightgrav

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    what did you use for the final velocity?
    ... by the way, this distance is NOT the height!
     
  12. Sep 30, 2007 #11
    Very good point. I hadn't realized that at first. So how do I go about finding the vertical height?
     
  13. Sep 30, 2007 #12

    learningphysics

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    use trig. When you get the distance along the ramp... the height is just dsin(38).
     
  14. Sep 30, 2007 #13
    How do I find the distance? By using Vxf^2 = Vxi^2 + 2(ax)(delta x)?
     
  15. Oct 1, 2007 #14
    I'm still confused about this problem.
     
  16. Oct 2, 2007 #15

    learningphysics

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    Yes. What did you get for distance?
     
  17. Oct 2, 2007 #16
    I got 1.459m for the distance
     
  18. Oct 2, 2007 #17

    learningphysics

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    How did you get that? Can you show your calculations?
     
  19. Oct 2, 2007 #18
    Well, I assumed that Vxf would be zero since the block would stop on the ramp. Then I used 14 for Vxi and -67.179 for the acceleration.

    -196 = 2(-67.179)deltax
    -196 = -134.358deltax
    delta x = 1.459
     
  20. Oct 2, 2007 #19

    learningphysics

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    -67 isn't the right acceleration. how did you get that?
     
  21. Oct 2, 2007 #20
    -(ukn)-Wx = m(ax)
     
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