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Another problem

  1. Nov 10, 2005 #1
    A 2 kg block is attached to the bottom of a spring that has spring constant of 50 N/m and block is at first at equilibrium point where it will hang motionless if undisturbed. Where will it be when it reaches its maximum speed of 4.0 meters per second?

    I thought it would be the equilibrium point because that is when it has the greatest force of gravity acting on it.
    Last edited: Nov 10, 2005
  2. jcsd
  3. Nov 10, 2005 #2
    The force of gravity is constant through out. Start with the position equation for simple harmonic motion.
  4. Nov 10, 2005 #3
    The gravitational field is calculated from g = -GM/r^2 where the negative sign indicates that the field is toward mass M, which in this case would be the mass of the Earth. G is the gravitational constant and r is the distance from the center of the object of mass M and the point you are interested in. Therefore the value of g increases as you move closer to the center of the Earth, so the gravitational force is at it's maximum when the block is at it's lowest point.

    You are however correct in saying the maximum speed occurs at the equilibrium point, and you can argue that this is true from energy considerations without doing any calculations.

    Edit: Forgot to mention, the distances that the block would move would be so small compared to the radius of the Earth that the force due to gravity may be treated as a constant.
    Last edited: Nov 11, 2005
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