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  1. #1
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    Mechanics Question

    A body of mass 5kg is attached to one end of a nonlinear spring which does not
    obey Hooke’s law. The other end of the spring is fixed, and no other forces act on the
    particle, which moves in one dimension. The force exerted by the spring is given in
    newtons by
    F = 8tan^-1(x/3)ex

    where x is the extension of the spring beyond its natural length measured in metres,
    and ex is a unit vector pointing in the direction of increasing x.
    The spring is extended by a distance of 4 m; the body is held at rest for a moment
    and then released. By considering the potential V (x) corresponding to the force F, or
    otherwise, determine the speed at which the body is moving when it reaches the point
    x = 0.
    [Hint: integration by parts may be useful in order to determine V (x).]

    ex is really e and a small x slightly below the e but i couldnt figure out how to insert here.
    I havent a clue where to start nor do any of my classmates! Help!
    Last edited by ly667; April 15th 2012 at 09:58 AM.
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  2. #2
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    Re: Mechanics Question

    To get started find x when mass hangs freely We know in this position F=5g So solve 5g=8tan^-1(x/3)
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  3. #3
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    Re: Mechanics Question

    Quote Originally Posted by ly667 View Post
    A body of mass 5kg is attached to one end of a nonlinear spring which does not
    obey Hooke’s law. The other end of the spring is fixed, and no other forces act on the
    particle, which moves in one dimension. The force exerted by the spring is given in
    newtons by
    F = 8tan^-1(x/3)ex

    where x is the extension of the spring beyond its natural length measured in metres,
    and ex is a unit vector pointing in the direction of increasing x.
    The spring is extended by a distance of 4 m; the body is held at rest for a moment
    and then released. By considering the potential V (x) corresponding to the force F, or
    otherwise, determine the speed at which the body is moving when it reaches the point
    x = 0.
    [Hint: integration by parts may be useful in order to determine V (x).]

    ex is really e and a small x slightly below the e but i couldnt figure out how to insert here.
    I havent a clue where to start nor do any of my classmates! Help!
    you sure the spring force equation is not F = -8 \arctan\left(\frac{x}{3}\right) ? (force should be opposite to displacement)

    \frac{dV}{dx} = -F

    V = 8 \int \arctan\left(\frac{x}{3}\right) \, dx

    integrating by parts ...

    u = \arctan\left(\frac{x}{3}\right) , dv = dx

    du = \frac{3}{9+x^2} \, dx , v = x

    8 \int \arctan\left(\frac{x}{3}\right) \, dx = 8\left[x \cdot \arctan\left(\frac{x}{3}\right) - \int \frac{3x}{9+x^2} \, dx \right]

    V = 8\left[x \cdot \arctan\left(\frac{x}{3}\right) - \frac{3}{2} \log(9+x^2) + C \right]

    since V(0) = 0 ...

    V = 8\left[x \cdot \arctan\left(\frac{x}{3}\right) - \frac{3}{2} \log(9+x^2) + \frac{3}{2} \log(9) \right]

    V = 8\left[x \cdot \arctan\left(\frac{x}{3}\right) + \frac{3}{2} \log \left(\frac{9}{9+x^2} \right) \right]

    set V(4) = \frac{1}{2} mv^2 to find the speed of the mass as it passes equilibrium
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  4. #4
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    Re: Mechanics Question

    I was taking it to be moving vertically so the forces were F upwards and 5g downwards. On reading it again I realize it's moving along an x axis
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  5. #5
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    Re: Mechanics Question

    No it is definitely just tan!
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    Re: Mechanics Question

    Quote Originally Posted by ly667 View Post
    No it is definitely just tan!
    huh? \tan^{-1}(x) = \arctan(x) , the inverse tangent function.
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  7. #7
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    Re: Mechanics Question

    Sorry I thought you meant in me typing out the question. What about the other value in that given equation, the ex?
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    Re: Mechanics Question

    Mechanics Question-mechanics-question.png

    That is the entire question, my lamda value is 4.
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  9. #9
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    Re: Mechanics Question

    Quote Originally Posted by ly667 View Post
    Sorry I thought you meant in me typing out the question. What about the other value in that given equation, the ex?
    e_x is a unit vector in the direction of the spring's displacement. The negative sign in front of the force expression makes the force vector opposite in direction to the given unit vector.
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