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Box sliding down a ramp question

  1. Jul 16, 2011 #1
    In order to get an appreciation for the time scales involved, think about a block that slides without friction down a ramp that is inclined at an angle q above the horizontal. Initially the block is at rest a distance s (measured along the plane) from the bottom. Click the button marked “DATA” for specific values of q and s. (Remember these numbers; you will need them later.) Calculate the time required for the block to slide to the bottom.

    For my given values I have: distance, s = 306 m, and q = 11 deg
    To find the time it takes for the block to slide to the bottom, I tried:

    d = 306sin(11) = 58.4 m

    d = v0t - 1/2at^2

    58.4 = 0 - 1/2(-9.8)t^2

    t = 3.45017

    and

    1/2mv^2 - 1/2mv0^2 = F*d

    1/2v^2 - 1/2(0)^2 = a*d

    1/2v^2 = -9.81(-306sin(11))

    v = 33.38289 m/s

    v = v0 + at

    t = 3.44898
     
  2. jcsd
  3. Jul 16, 2011 #2
    I can see how you tried this problem. However, when a block is on a ramp the situation is slightly different. Take a look at the attached diagram. This is how the forces are acting on this block. The force normal comes into play, and of course in this case, there is no force friction. So the only two forces in action are the weight of the block and the force normal.

    Take the plane of the ramp as the x-axis, as on the diagram. Then the force normal is a y-force, the weight has x and y components. The y-component and force normal should cancel out (can you see why?) and we are left with the x-component of the weight, which is the reason the block moves. Equate that with ma, find the acceleration, then go on as you did in your solution.
    Hope this helps, let me know if you have any questions! :)
     
    Last edited: Jul 16, 2011
  4. Jul 16, 2011 #3

    PhanthomJay

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    This is the height of the block at its start point, not the distance travelled by the block
    you have to correct the distance and also the accelerataion. The acceleration is not g, the block is not in free fall. Newton's laws will help you find the acceleration down the plane.
    this looks OK
    Correct the value of 'a' using the value you found in part 1.
     
  5. Jul 16, 2011 #4
  6. Jul 16, 2011 #5
    whoops... arithmetical error

    once I fixed it I got

    t = sqrt(306/.5*9.8*sin(11)) = 18.901, which was correct

    thank you for the help :)
     
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