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!

Nonlinear spring energy problem

  1. May 2, 2009 #1

    lzh

    User Avatar

    1. The problem statement, all variables and given/known data
    The stretch of a nonlinear spring by an amount x requires a force F given by:
    F=40x-6x^2
    where F is in Newtons and x is in meters.

    What is the change in potential energy U when the spring is stretched 2m from its equilibrium position?


    2. Relevant equations
    U=.5kx^2
    = .5(kx)x
    F=kx

    3. The attempt at a solution
    F=40(2)-6(2)^2=80-24=56N
    U=.5(56N)(2)=56J

    Thats what I thought would work, but 56J is no the correct answer. What am I doing wrong?
     
  2. jcsd
  3. May 2, 2009 #2
    F=-dU/dx

    use this
     
  4. May 2, 2009 #3

    lzh

    User Avatar

    So it'd be:
    56N(2m)=-dU
    dU=-112?
    Am I not getting this?
     
  5. May 2, 2009 #4

    Astronuc

    User Avatar

    Staff: Mentor

    The trick here is to recognize that F=40x-6x^2 which is not the same as F=kx.

    Now U(x) = [tex]\int^x_0\,F(x) dx\,=\,\int^x_0\,kx\,dx[/tex]

    http://hyperphysics.phy-astr.gsu.edu/hbase/pespr.html#pe2

    http://hyperphysics.phy-astr.gsu.edu/hbase/pegrav.html#pe
     
  6. May 2, 2009 #5

    lzh

    User Avatar

    Oh, I see!

    Thanks, I got it!
     
  7. May 2, 2009 #6
    f=-dU/dx
    integrating both sides
    20x^2 -2x^3=-U
    put x=2
    U=-64J

    change would be 64...and work done would be 64 J
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Nonlinear spring energy problem
Loading...