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Potential energy of a particle in a system

  1. Oct 8, 2014 #1
    • Warning! Posting template must be used for homework questions.
    1. A single conservative force acting on a particle within a system varies as
    Farrowbold.gif = (− Ax + Bx6)ihat N, where A and B are constants, Farrowbold.gif is in newtons, and x is in meters.
    (a) Calculate the potential energy function U(x) associated with this force, taking U = 0 at x = 0.
    (b) Find the change in potential energy and change in kinetic energy as the particle moves from
    x = 1.30 m to x = 3.60 m.

    2. I got Ax^2/2 - b^7/2 but it was wrong so I'm really confused on how to go about this problem

    Thanks!!
     
    Last edited: Oct 8, 2014
  2. jcsd
  3. Oct 8, 2014 #2

    HallsofIvy

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    The potential energy corresponding to force [itex]\vec{F}[/itex] is a scalar function [itex]\phi(x,y)[/itex] such that [itex]\nabla \phi= -\vec{F}[/itex]. Of course, [itex]\nabla \phi[/itex] is defined as [itex]\frac{\partial \phi}{\partial x}\vec{i}+ \frac{\partial \phi}{\partial y}\vec{j}[/itex]. But it is not clear to me what "F" is- is "N" a vector? If so, what vector?

    If you mean that [itex]\vec{F}= (-Ax+ Bx^6)\vec{i}[/itex] then we must have
    [tex]\frac{\partial \phi}{\partial x}= Ax- Bx^6[/tex]
    [tex]\frac{\partial \phi}{\partial y}= 0[/tex].

    From [itex]\partial \phi/\partial x= Ax- Bx^6[/itex], we have
    [tex]\phi(x,y)= \frac{A}{2}x^2- \frac{B}{7}x^7+ p(y)[/tex]
    But then
    [tex]\frac{\partial \phi}{\partial y}= p'(y)= 0[/tex]
    so that p(y) is actually a constant:
    [tex]\phi(x, y)= \frac{A}{2}x^2- \frac{B}{7}x^7+ C[/tex]
     
    Last edited: Oct 8, 2014
  4. Oct 8, 2014 #3
    Hi, thanks for the help! that was the same answer I got but it was wrong and I don't really understand what I'm supposed to do
     
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