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: Integral of (Force * Velocity).

  1. Oct 7, 2008 #1
    1. The problem statement, all variables and given/known data

    A particle moves in the x-y plane having the components of its velocity to be:

    [tex]x = 64\sqrt{3}t[/tex] and [tex]y = 64t - 16t^2[/tex],

    and a force acting on this particle is proportional to its velocity. Find:

    [tex]\int(F \cdot V)dt[/tex]

    from t = 0 to t = 4. Give a physical meaning to your result.



    2. Relevant equations

    Not sure.



    3. The attempt at a solution

    I'm having a hard time getting started here, because I don't know what F is. I've got:

    [tex]V = (64\sqrt{3}t)i + (64t - 16t^2)j[/tex],

    right? But I don't know what to dot it with inside the integral. I'm not looking for a total solution here, I'm just wondering if someone can quickly tell me what exactly F is. I should be ok from there.

    If F is proportional to V, do I just set

    [tex]F = (64a\sqrt{3}t)i + (64at - 16at^2)j[/tex]

    for some unknown constant a?

    Thanks.
     
    Last edited: Oct 7, 2008
  2. jcsd
  3. Oct 7, 2008 #2
    So, if I set F equal to what I stated above, I get an answer of:

    [tex]\int( F \cdot V ) dt = \int(16384 a t^2 - 2048 a t^3 + 256 a t^4)dt = (4063232 a)/15[/tex]

    I don't even know what to make of that...
     
    Last edited: Oct 7, 2008
  4. Oct 7, 2008 #3
    Why would the force be proportional to the velocity?
    You know (I hope) that F=m*a.
    And acceleration is the derivative of velocity, isn't it?

    The significance of the integral... v*dt is ds (distance). Force time distance = ???
     
  5. Nov 25, 2010 #4
    Force times Distance equals Work
     
  6. Nov 25, 2010 #5

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Because it's a given in the problem.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook