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[Work/Energy] a particle inside a field

  1. Dec 23, 2011 #1
    1. The problem statement, all variables and given/known data

    a particle is moving in a field where it's potential energy is [itex]U= -kx+c[/itex] , c is a constant

    it is also experiencing a damping force [itex]F=-b\vec{v}[/itex]
    it has an initial velocity [itex]v=V_0\hat{y}[/itex]

    d. what is the work done by the net force throughout all of the particle's movment?
    e. what is the power of the damping force when the acceleration of the particle is 0?


    2. Relevant equations

    in first the rest of the questions I have found :
    the velocity:
    [itex] v_x=-\frac{k}{b} (1-e^{-{frac{b}{m}t})[/itex]
    [itex] v_y= V_0 e^{-\frac{b}{m}t} \hat{y}[/itex]

    the velocity when a=0:
    [itex]v_max = -\frac{k}{b}\hat{x}[/itex]

    the net force
    [itex]F=ke^{-\frac{b}{m}t}\hat{x} -bV_0e^{-\frac{b}{m}t}\hat{x}[/itex]

    3. The attempt at a solution
    d. on one hand the force acts "until" [itex]t=\infty[/itex] so it should be infinity...
    on the other hand the force will be very small so I'm positive it isn't infinity :/

    e. the damping force at a=0 or [itex]t=\infty[/itex] will be quite obviously
    [itex]F=k\hat{x}[/itex]
    (that way it will be equal in magnitude and opposite in direction of the force causing the particle to move)

    but we need the power so it's basically
    [itex]P=Work per second=F\cdot \vec{v_max}[/itex]
    which is
    [itex]P=\frac{k^2}{b}[/itex]
    am I wrong?
    I used the velocity because it's how much x is changed per second - it ok by units, but is it right?
    Thanks!
     
    Last edited: Dec 23, 2011
  2. jcsd
  3. Dec 24, 2011 #2
    you could use the work-energy theorem, or just integrate the net force on the path of the particle..
     
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