1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook