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Horizontal Distance up Incline

  1. Oct 11, 2015 #1
    1. The problem statement, all variables and given/known data
    A spring with a spring constant = 1000 N/m is compressed 0.20 m then launches a block of mass 200 g. The horizontal surface is frictionless and the coefficient of kinetic friction with the inline is 0.15.
    What horizontal distance does the block cover while in the air after it takes off at the top of the incline?
    (diagram attached)

    2. Relevant equations


    3. The attempt at a solution
    deltaK + deltaUg + deltaUs = 0
    (1/2mv^2f -1/2mv^2i) + 0 + (1/2kx^2f - 1/2kx^2i) = 0
    1/2mv^2f = 1/2kx^2i

    v = sqrt(kx^2/m)
    v = 14.14 m/s

    Distance up incline:

    x = 2.5m/sin50
    =3.264 m

    Acceleration:
    -mgsin0 - ukmgcos0 = ma
    a = -gsin50-(0.15)(-9.81)cos50
    a = 6.562


    Speed at the top:
    v^2 = vi^2 +2ad
    v = sqrt[14.14^2 + 2(6.562)(3.264)]
    v = 15.58 m/s

    I'm not sure what the next step would be (also not sure if the above steps are correct).
    To find horizontal distance covered would I have to find the horizontal time and then using speed = distance/time solve for distance?

    Thanks =]
     

    Attached Files:

  2. jcsd
  3. Oct 11, 2015 #2
    How can the speed at the top be greater than the speed at the bottom?
     
  4. Oct 11, 2015 #3
    Oops I think I got some of the signs wrong

    Acceleration:
    -mgsin0 - ukmgcos0 = ma
    a = -gsin50-(0.15)(9.81)cos50
    a = -8.456 m/s^2

    so velocity at the top:
    v^2 = vi^2 +2ad
    v = 14.14^2 + 2(-8.452)(3.264)
    v = 12.03 m/s
     
  5. Oct 11, 2015 #4
    OK, much better.

    Your idea to find the horizontal distance might not work as you stated it. Can you try something else along the same idea?
     
  6. Oct 11, 2015 #5
    At max height v = 0
    Time to reach max height:
    t = visin0/g
    t = 12.03sin50/9.81
    t = 0.9394 seconds

    x= xo + vit
    x = 0 + (12.03)(0.9394)
    x = 11.3 m
     
  7. Oct 12, 2015 #6
    Good try but you made two mistakes:
    1) The time t you calculated is only the time for the block to reach the maximum height. What about the time it takes to come back down?
    2) What direction is vi going in?
     
  8. Oct 12, 2015 #7
    Would I have to find the time of flight for the whole range of the parabola first?
    time of flight:
    y = yo + 1/2gt^2
    2.5 = 0 + 1/2(9.81)t^2
    t = 0.7143 s

    Horizontal distance:
    x= xo + vit
    x = 0 + (12.03)(0.7143)
    x = 8.6 m
     
  9. Oct 12, 2015 #8
    No, you break it into separate x and y components. You started off right in your first attempt but didn't finish it for some reason.
     
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