1. Not finding help here? Sign up for a free 30min 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!

Force and integration

Tags:
  1. Feb 18, 2017 #1
    Hello! First time poster, please treat me well! :wink: I've already solved the problem below on my second attempt with the help of kinetic energy but I want to know why my first attempt gives a wrong answer.

    1. The problem statement, all variables and given/known data

    A force in the +x-direction with magnitude F(x) = 18.0 N - (0.530 N/m)x is applied to a 6.00-kg box that is sitting on the horizontal, frictionless surface of a frozen lake. F(x) is the only horizontal force on the box. If the box is initially at rest at x = 0, what is its speed after it has traveled 14.0 m?

    2. Relevant equations
    a = F/m

    3. The attempt at a solution
    F(x) = 18.0 - 0.530x =>
    a(x) = 18.0/m - 0.530x/m =>
    Integrate =>
    v(x) = 18.0x/m - 0.530x2/2m = 3x - 0.530/12 x2
    v(14) = 33.3

    Why is this incorrect? I'm guessing that the integral of a(t) is velocity but the integral of a(x) is something else.
     
    Last edited: Feb 18, 2017
  2. jcsd
  3. Feb 18, 2017 #2

    berkeman

    User Avatar

    Staff: Mentor

    Welcome to the PF :smile:

    When you integrate x, you don't just get x^2... :smile:
     
  4. Feb 18, 2017 #3
    Oops, I edited my question. The result is still wrong though.
     
  5. Feb 18, 2017 #4

    berkeman

    User Avatar

    Staff: Mentor

    Looks like it's just a math error when you plug in x=14. Try again?
     
  6. Feb 18, 2017 #5
    I typed this into wolfram alpha "3*14 - 0.53/12*14^2" and got 33.3. When I use W = K2 - K1, W = F*s with the average force from x = 0 to x = 14 I get the velocity at 14 meters to be 8.17 which is the correct answer.
     
  7. Feb 18, 2017 #6

    Doc Al

    User Avatar

    Staff: Mentor

    Exactly. If you integrate F(x)dx, you can calculate the work and thus the change in KE.
     
  8. Feb 18, 2017 #7

    berkeman

    User Avatar

    Staff: Mentor

    You should use parenthesis, I believe. x^2 is not in the denominator...
     
  9. Feb 18, 2017 #8

    berkeman

    User Avatar

    Staff: Mentor

    Ah, so the choice of integral is wrong, regardless of errors in the integration. Thanks Doc.
     
  10. Feb 18, 2017 #9
    Ahh, so if F(x)dx integrated is work I'm guessing that integrating a unit1 with unit2 equals unit1*unit2. So when I integrated a(x)dx I actually got meter2/time2 instead of velocity. Is this correct?
     
  11. Feb 18, 2017 #10

    Doc Al

    User Avatar

    Staff: Mentor

    That's true.
     
  12. Feb 18, 2017 #11

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    It is easy to get confused here.
    It depends whether you are thinking of a as the physical entity acceleration or as purely a mathematical function.
    As a physical entity, it's not whether you think of a as a function of time or of displacement that matters. What matters is what you integrate with respect to, as you figured out. Integration wrt a variable is akin to multiplication by that variable. ∫a.dt gives Δv, while ∫a.dx gives Δ(v2)/2.
    If you are given a as a function of displacement, ∫a.dt still yields velocity, but how are you to perform the integral? E.g. if told a=kx3, your integral is ∫kx3.dt. Since you do not know what x is as a function of t this is not directly solvable. If you are told the total displacement then you can use ∫a.dx instead to find the velocity, but what if you are only told the elapsed time? In that case you would have to write out and solve the differential equation.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Force and integration
  1. Integrating Forces (Replies: 4)

  2. Impulse force integral (Replies: 2)

Loading...