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!

Work-Kinetic Energy question

  1. Sep 18, 2011 #1
    This question is whooping me. I know I have the calculus part right I just dont know where to go from here?

    A particle moving along the x axis is acted upon by a single force F = F0e^(–kx), where F0 and k are constants. The particle is released from rest at x = 0. It will attain a maximum kinetic energy of

    F0/k
    F0/e^(k)
    kF0
    1/2(kF0)^2
    ke^(k)F0

    So using the Work Kinetic Energy theorem
    W=Kf-Ki since x=0 then Kf=W

    W=integral(F = F0e^(–kx),x,0,xf)
    taking F0 out, since its a constant, and integrating the exponential function I get
    W=F0[-e^(-kxf)/k + 1/k]

    now what????
     
  2. jcsd
  3. Sep 18, 2011 #2
    When you evaluate the integral one of the terms is zero,

    exp(-x) goes to zero as x goes to infinity?
     
  4. Sep 20, 2011 #3
    I did that, that's why the second term inside the brackets is 1/k, since e^(-k*0)=1.
     
  5. Sep 20, 2011 #4

    BruceW

    User Avatar
    Homework Helper

    You've pretty much done it now. So you now have 1/k inside the bracket and f0 outside the bracket, so just multiply them together to get the answer.
     
  6. Sep 24, 2011 #5
    Ok, that makes sense. I was there I just couldn't see it. Thanks all.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Work-Kinetic Energy question
Loading...