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: Position dependant forces

  1. Oct 12, 2004 #1
    the question asks,
    Find the velocity v as a function of the displacement x for a particle of mass m, which starts from rest x=0
    F(x)=Fo + Cx where Fo and C are positive constants
    So far i've gotten,
    ma=Fo + Cx
    m (dv/dt)=Fo +Cx
    m (dv/dx dx/dt)=Fo +Cx I split dv/dt using the product rule
    m v dv=(Fo + Cx) dx v=dx/dt
    now i'm haveing problems doing the intergal of both sides i have so far
    m(v-vo)=????

    can anyone help?
    I've intergrated on the LHS from vo to v and i think the RHS should be from xo to x
     
  2. jcsd
  3. Oct 12, 2004 #2

    Pyrrhus

    User Avatar
    Homework Helper

    What the problem?

    [tex] mvdv=(F_{o} + C_{x})dx [/tex]

    [tex] \int_{v_{o}}^{v} mvdv= \int_{x_{o}}^{x} (F_{o} + Cx)dx [/tex]

    [tex] \int_{v_{o}}^{v} mvdv= \int_{x_{o}}^{x} F_{o}dx + \int_{x_{o}}^{x} Cxdx [/tex]

    Mass is a constant
    Fo and C are positive constants

    Integrate.

    [tex] m \int_{v_{o}}^{v} vdv= F_{o} \int_{x_{o}}^{x} dx + C \int_{x_{o}}^{x} xdx [/tex]

    You should have:

    [tex] m \frac{1}{2}v^2]_{v_{o}}^{v}= F_{o}x]_{x_{o}}^{x}+ \frac{1}{2}Cx^2]_{x_{o}}^{x} [/tex]

    [tex] \frac{1}{2}m(v^2 - v_{o}^2)= F_{o}(x-x_{o})+ \frac{1}{2}C(x^2-x_{o}^2) [/tex]

    Our Particle starts from rest (Vo = 0) at x = 0 so

    We got

    [tex] \frac{1}{2}m(v^2)= F_{o}x+ \frac{1}{2}Cx^2 [/tex]

    [tex] v(x) = \sqrt{\frac{2F_{o}x+ Cx^2}{m}} [/tex]

    There you go.
     
    Last edited: Oct 12, 2004
  4. Oct 12, 2004 #3
    [tex] m \int_{v_{o}}^{v} vdv= F_{o} \int_{x_{o}}^{x} dx + C_{x} \int_{x_{o}}^{x} dx [/tex]
    should be

    [tex] m \int_{v_{o}}^{v} vdv= F_{o} \int_{x_{o}}^{x} dx + C \int_{x_{o}}^{x} xdx [/tex]
    sorry i didn't explain it well enough C and X are seperate

    so does that mean the solution would be:
    [tex] 1/2mv^2=F_{o}x + 1/2Cx^2 [/tex]
    [tex] v= \sqrt {(2F_{o}x)/m + (Cx^2)/m} [/tex]
     
  5. Oct 12, 2004 #4

    Pyrrhus

    User Avatar
    Homework Helper

    Oh ok no problem, but at least you got it :smile:
     
  6. Oct 12, 2004 #5
    [tex] v_{o} = 0 , x_{o} = 0 [/tex]


    thanks for your help
     
  7. Oct 12, 2004 #6

    Pyrrhus

    User Avatar
    Homework Helper

    Always a pleasure to help, and Welcome to PF! :smile:
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook