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    The force acting on a body of mass 5kg is inversely proportional to the square of the distance the body is from an origin, O, and is directed away from O. When the body is 6m from O and moving away from O, in the direction of the force, with speed 10m/s, the force of the body is 30N.
    a) show that v^2 = 172 - 432/s
    where v m/s is the speed of the body when it is s m from O.
    b) determine the speed the body approaches, but does not exceed, as it moves further and further from O.

    Have fun?

    F= k/(s^2) = 5*(dv/dt)

    s=6, v=10, F=30 gives k=1080.

    So 1080/(s^2) = 5*(dv/dt) = 5*(dv/ds)*(ds/dt) = 5*(dv/ds)*v
    So 216/(s^2) = 5v*(dv/ds)
    Separate the variables and solve!

    For part b) what does v^2 approach as s tends to infinity?

    Start with \dfrac{d^2 x}{dt}=\dfrac12 \dfrac d{dx} v^2 . Express the left as Force/mass. Then integrate (remembering the constant).

    a.)F = \frac{k}{x^2}

    ma = \frac{k}{x^2}

    mv\frac{dv}{dx} = \frac{k}{x^2}

    \frac{1}{2}mv^2 = \frac{-k}{x} + C

    Conditions: (m=5kg, v=10ms, x=6m, F=30N)

     30 = \frac{k}{6^2}

     k = 1080

     \frac{5}{2}(10)^2 = \frac{-1080}{6} + C

     C = 430

    Therefore: \frac{5}{2}v^2 = \frac{-1080}{x} + 430

     v^2 = \frac{-432}{x} + 172

    b.)  v = \sqrt{172-\frac{423}{x}}

    As "x" tends to infinity, equation reduces to:

     v = \sqrt{172}

    I don't care about the rep, I enjoyed myself by doing that :sexface: (Probably wrong... :p:)
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Updated: April 5, 2011
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