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Math Help - Energy and equation of motion of springs

  1. #1
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    Energy and equation of motion of springs

    Hi All

    I have one precious hour before this needs to be in and ive spent all night on it to no avail. Please help?

    For particle P how do I calculate the gravitational potential and kinetic energy at a general point of its motion?

    Also, what is the potential energy stored in each spring at the same instant and how do i write down an equation representing the conservation of mechanical energy E for the system?

    Please please pretty please help
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    Am i right in thinking that the potential energy is in the direction of the springs? i.e. either side of the particle - one going up the slope and the other coming down?

    What about the kinetic energy?
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    Forum Admin topsquark's Avatar
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    I'll get back to you later on this if someone else doesn't do it first.

    -Dan
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    Hi Topsquark

    Can you help please?

    Is the potential energy found by using E=0.5mv^2+mgx and appliying it at the suitable angle to each force?

    e.g U(x)=2mgxsin(theta)
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by moolimanj View Post
    Hi All

    I have one precious hour before this needs to be in and ive spent all night on it to no avail. Please help?

    For particle P how do I calculate the gravitational potential and kinetic energy at a general point of its motion?

    Also, what is the potential energy stored in each spring at the same instant and how do i write down an equation representing the conservation of mechanical energy E for the system?

    Please please pretty please help
    The gravitational potential energy is easy. I'd pick the point at A to be the zero level, so
    [tex]U = mgh = mgx~sin(\alpha)[tex]

    As to the kinetic energy, the KE of an object on a spring with motion equation y = A~cos( \omega t) is given by
    K = \frac{1}{2}mv^2 = \frac{1}{2}m \omega ^2 A^2~sin^2( \omega t)

    In this case we have two springs, both fitting that form for the motion. (Though the top spring is actually the form y = -A~cos( \omega t).) So I get for the bottom spring:
    K =  \frac{1}{2}m \left ( \frac{2k}{m} \right ) A^2~sin^2 \left ( t \sqrt{\frac{2k}{m}} \right )
    where A is the amount the spring has been stretched from its equilibrium length. You mentioned this information in an earlier post, but I don't recall what it was.

    For the top spring we have a similar equation:
    K =  \frac{1}{2}m \left ( \frac{3k}{m} \right ) A^2~sin^2 \left ( t \sqrt{\frac{3k}{m}} \right )
    where a similar comment to the above is made for A.

    The total mechanical energy is, of course, the sum of these three energies.

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by moolimanj View Post
    Am i right in thinking that the potential energy is in the direction of the springs? i.e. either side of the particle - one going up the slope and the other coming down?

    What about the kinetic energy?
    Be careful about this: energy is a scalar quantity; it has no direction. So we can't talk about a potential energy being in "the direction of the springs."

    -Dan
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