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Block on incline with air resistance, no friction

  1. Sep 4, 2013 #1
    1. The problem statement, all variables and given/known data

    A textbook of mass m = 1.24 kg starts at rest on a
    frictionless inclined plane (angle  = 30◦). Although there is
    no friction, suppose there is a drag force (due to air resistance)
    acting on the book which is proportional to the speed squared
    and is described by the equation F = kmv2, where k = 0.86
    m−1. How much time does it take for the textbook to slide a
    distance d = 1.65 m down the plane? (Hint: This one is tricky,
    you will need to solve the integral by hand using a hyperbolic
    trig substitution.)

    2. Relevant equations

    Just F=m*a
    drag force = k*m*v^2

    3. The attempt at a solution

    So the mass keeps canceling on me whenever I run the problem, don't know if I actually need it for this one. I changed my coordinate system so positive x was parallel to the incline and positive y was perpendicular to it.

    Forces are gravity broken in componets now, normal force only acting in positive y, canceled out by gravities y component. drag force in -x direction and +x component of gravity.
    Vf= final velocity
    y- direction F=ma
    F= 0, its moving down the incline

    x - direction F= ma
    m*g*cos(theata)-k*m*v^2 = m dv/dt
    solve it for time with given conditions and I got
    T = (1/(Sqrt[g*sin(theata)*k])*ArcTanh[Sqrt[k/(g*sin(theata))]*Vf]

    Switch the beginning equation to m*g*cos(theata)-k*m*v^2 = m*v* dv/dx and solve for the distance given the distance yields:
    x = (1/2*k)*Ln((g*sin(theata)-k*Vf^2)/(g*sin(theata)))
    Solve that equation for Vf yields:
    Vf = Sqrt[(g*sin(theata)/k)*(1-e^(2*k*x))/k)
    Sub into T equation and get:
    T = (1/(Sqrt[g*sin(theata)*k])*ArcTanh(1-e^(2*k*x))

    I keep getting very small times less than one and complex numbers which I feel are wrong, well its telling me they are wrong. I might have set it up wrong when defining my coordinate system or something. if someone can help me that would be awesome, even if its on a regular defined coordinate axes.
  2. jcsd
  3. Sep 14, 2013 #2
    It's an old post so I'm not sure if you still want advice, but I think the issue comes with your second diff eq you set up to solve for the distance.

    You can save a lot of work and simply turn your first diff eq into a function of distance right off the bat by using derivatives of ##x(t)## in place of ##v## and ##a##. This can be integrated without trouble given 2 initial conditions on velocity and position. From there, you should have a function ##x(t)## that you can use to plug in your numbers and numerically invert. It seemed to give a reasonable real value for time when I worked it out.
  4. Sep 14, 2013 #3


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    Homework Helper
    Gold Member

    You miss a negative sign in front of the right-hand side.

    The data might be wrong. The speed can not exceed √(gsinθ/k), but the value obtained with the given d and k does.
    Last edited: Sep 14, 2013
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