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Conservation of Energy Problem

  1. Oct 21, 2009 #1
    1. The problem statement, all variables and given/known data
    A particle can only move along the x axis. Forces act on it so that its potential energy function is U(x) = 1/2*k1*x^2 + 1/4*k2*x^4 where k1 and k2 are positive. The particle is started at x = a with zero velocity.
    a.) Where is the velocity a maximum? What is its magnitude?
    b.) Where else will the velocity be zero?
    c.) What is the force on the particle as a function of x?


    2. Relevant equations
    F = d/dx -U(x)
    W = -U(x2) + U(x1)


    3. The attempt at a solution
    It's asking for the point where the velocity is maximum so the acceleration has to equal zero or not exist. The function exists at all x-values so we need to find the acceleration as a function of x.

    since F = d/dx -U(x), F = -(k1*x + k2*x^3)

    F = ma

    ma = -(k1*x + k2*x^3)
    a = -1/m*(k1*x + k2*x^3)
    When a = 0
    0 = -1/m*(k1*x + k2*x^3)
    -k1*x = k2*x^3
    x^2 = -k1/k2
    x = sqrt(-k1/k2)

    I'm completely lost at this point. Thanks for the help!
     
  2. jcsd
  3. Oct 21, 2009 #2

    kuruman

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    You can't say x=sqrt(-k1/k2) for two good reasons (a) x is an independent variable - it tells you where the particle is - so it cannot be constant and (b) the negative of a square root has no physical meaning whereas x denotes where the particle is at a given time.

    Look at your expression again,

    0 = -1/m*(k1*x + k2*x^3)

    For what value(s) of x is the right side zero?

    For part (b): You know that at x = a the kinetic energy is zero and the potential energy is
    U = (1/2)k1a2+(1/4)k2a4. Can you find another value for x, other x = +a where the potential energy has the same value? If you can, then you know that by conservation of energy the kinetic energy will be zero there as well.
     
    Last edited: Oct 21, 2009
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