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Mass, spring, inclined plane

  1. Oct 21, 2005 #1
    I have a question about thse following system: block of mass M is attached to a spring is on an inclined plane.
    Why do the force and energy analysis equations give different answers? The question is to find the maximum extension of the spring. Here is my thinking so far.

    I know if you use energy convervation, you get 1/2*k*x^2 = M*g*sin(theta), so you get x = 2*M*g*sin(theta)/k.
    However, why can I not just set the force along the incline equal to the spring force and solve for x that way? If I do that, I get a different answer:
    k*x = M*g*sin(theta)
    x = M*g*sin(theta)/k and this is different by a factor of 2.
    Last edited: Oct 21, 2005
  2. jcsd
  3. Oct 21, 2005 #2


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    What exactly is the situation here? Could you draw a diagram?
    What's with the mgsin(theta)? It's a force, not work. You cannot use it in the energy conservation equation.
  4. Oct 22, 2005 #3


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    The problem is the energy convervation. The potential energy should be
    M*g*h were h is the hight difference of the two states (before the mass was released and after).
    What about kinetic energy as the weight get to the end? (it has mass and it is moving)
    Also, the way that you did the first answer is missing a square route.
    If you can find the connection between the h and the x then the ^2 could cancel out!
  5. Oct 24, 2005 #4


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    In the answer above I assumed that you wanted the equilibrium. If you want the maximum extension the you can't say that the force of gravity and the spring along the plane are equal because as the mass is accelerating as it gets to the bottom (harmonic motion)
  6. Oct 24, 2005 #5
    A spring with constant k, and a displacement (x) from static deflection position (dst), also called equilibrium position.

    Force: k*dst = W_block = mg
    mg - k(x+dst) = ma
    ma + kx = 0

    ma here gives the force on a spring that expands it by x from equilibrium position, dst. F = -kx, by convention direction of x is negative to the dst, so F = kx

    When you compress the spring you are doing mechanical work on it W = (F*dr) and storing energy (KE, PE, or both) in the spring. When the force is removed the spring does work on the block to move it a distance d against external forces. When it is not compressed it has no stored energy and a force is applied to expand the spring from its equilibrium position.
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