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Classical block on inclined plane

  1. Sep 12, 2013 #1
    1. The problem statement, all variables and given/known data

    A block at t=0 is at the bottom of the plane and is projected up an inclined plane with initial speed v0. The plane's acute angle is θ above the horizontal, and the coefficient of friction is μ between the block and plane.

    Find the time (t) and velocity (v1) of the block when it reaches a given distance d.

    2. Relevant equations


    1. F=ma
    2. d=(1/2)at^2 + v0t + d0
    3. v1=at+v0

    3. The attempt at a solution

    I drew a picture electronically:
    BVTIW9D.jpg

    I made the direction FN and v0 positive.

    #1
    x dir
    F = ma
    F = -FμN - mgsinθ
    ma = -μmgcosθ - mgsinθ
    a = g(-μcosθ - sinθ)

    So I found my a (should be a negative value), and now I want to find t when the block reaches d.

    #2
    d = (1/2)(-at^2) +v0t +d0
    0 = (1/2)(-at^2) +v0t + 0 - d


    Problem: Acceleration and distance are negative so the roots are imaginary, but I need to find real roots of t and then plug it in v1=at+v0 (I think).
     
  2. jcsd
  3. Sep 12, 2013 #2
    Why do you have -a in equation #2?

    And why do you think that the roots are imaginary? They depend on d; if it is sufficiently small, roots will be real. This makes perfect sense physically, because there is some maximum distance that the block can travel at any initial velocity.
     
  4. Sep 12, 2013 #3

    haruspex

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    That does not automatically make the roots imaginary. There will be real roots if v0 is nonzero and d is sufficiently small.
     
  5. Sep 13, 2013 #4
    It's just that when I graph 0 = (1/2)(-at^2) +v0t + 0 - d, the entire curve is below the x-axis.
     
  6. Sep 14, 2013 #5
    Consider ## a = 2, \ v_0 = 5, \ d = 4 ##: is your statement true?
     
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