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Length of a ramp

  1. Nov 13, 2014 #1
    1. The problem statement, all variables and given/known data
    You are passing a construction site on the way to physics class, and stop to watch for a while. The construction workers appear to be going on a coffee break, and have left a large concrete block resting at the top of a wooden ramp connecting one of the building's upper floors to the ground. As soon as their backs are turned, the block begins to slide and takes about 2 seconds to reach the ground. You wonder how long the ramp is. You estimate the ramp is at an angle of about 37* to the horizontal, and your physics book gives the coefficient of kinetic friction (uk) between concrete and wood as 0.25

    2. Relevant equations
    a= F/m
    vf=vi+at
    vf^2=vi^2+2ax

    3. The attempt at a solution
    I found that:
    Fnety = Fn - Fgy = Fn - mgsin(37) = 0
    Fn = mgsin(37)
    Fnetx = Ff - Fgx = uk(mg)sin(37) - mgcos(37) = ma
    a = uk(g)sin(37) - gcos(37)
    a = .25(-9.8)sin(37) + 9.8cos(37)
    a = 6.35 m/s^2

    vf-vi = at
    vf = 6.35(2)
    vf = 12.7 m/s

    vf^2 - vi^2 = 2ax
    12.7^2 = 2(6.35)x
    x = 12.7m

    I don't understand how this is wrong?
     
  2. jcsd
  3. Nov 13, 2014 #2

    Simon Bridge

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    Did you draw a free-body diagram?
    Check your trig relations - the angle is measured from the horizontal.
    Don't forget to define the x and y directions as part of your working.
     
  4. Nov 13, 2014 #3
    I did draw a free body diagram and I believe my trig relations are correct.
    After further digging I found nearly the exact same question here but when I calculate the velocity with their numbers I still get the wrong velocity?
    Edit: As in their paper says vf = 2.5 m/s but I get 80.36 when I plug those exact same numbers into my calculator. Did they do it wrong or am I messing up my calculation somewhere?
     
  5. Nov 14, 2014 #4

    NascentOxygen

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    I think you have mixed up your sines and cosines. You need a large, clear diagram to work on, or you probably will get these wrong.

    Is the correct answer around 7.5 roughly?
     
  6. Nov 16, 2014 #5

    Simon Bridge

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    Looking at your FBD: if you have theta as the angle of the ramp to the horizontal, and you have defined the +y axis to be upwards and perpendicular to the ramp, then the normal force to the ramp is ##\vec N = \hat\jmath mg\cos\theta##. You may need to redraw your diagram to see this.
     
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