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Simple Velocity Problem

  1. Sep 2, 2004 #1
    hi, i'm wondering if anyone could explain to me why one uses the following technique to solve this problem.

    Let's say we've a 10 kg object. The force exerted on it is

    F = 10 + 2t

    The force acts on it for two (2) seconds. If it starts at rest, what is the final velocity?

    I used this technique to solve the problem, after first making a mistake

    dp = F(net) * dt

    breaking that down to

    dp = (10 + 2t) * dt

    integration giving

    10t + t^2 from 0 to 2

    giving

    24 kg * m/s

    so

    dp = 24 kg * m/s
    m(v(2)-v(1)) = 24 kg * m/s
    10 kg (v(2)-0) = 24 kg * m/s
    v(2) = 2.4 m/s

    this is the correct answer, however, initially i tried

    m(v(2)-v(1)) = (10 + 2t) * dt

    10 kg (v(2)-0) = 14N * 2s

    resulting in a velocity of 2.8 m/s

    why do i use integration before involving momentum is my question. thanks for anyone who might be able to help!
     
  2. jcsd
  3. Sep 2, 2004 #2
    Even though i haven't taken calculus yet, according to your equation of the Force applied

    F = 10 + 2t

    The force applied during those 2 seconds is changing. thus you can't just use the quation F*t = change in momentum since you're saying the force of 14 N is applied for 2 seconds even though that wasn't true.
     
  4. Sep 2, 2004 #3
    i hadn't though of it in that manner. i suppose i should have since force is related to time. the only reason i thought to use integration is because i had the equation p = (10 + 2t) *dt. the area of the graph would be in units of momentum.

    thanks!

    (still would appreciate explanation of usage of integration other than realizing area of function is in units of momentum)
     
  5. Sep 2, 2004 #4
    Integrating a function is a method of finding the area beneath it. There's an elegant proof found in most calculus books that connects Rieman's sums (which is really just adding a bunch of infinately small shapes together to find area) to integration. Thus, since integration is much simpler than using Rieman's sums, integration is the preferred technique for finding area beneath a curve.
     
  6. Sep 2, 2004 #5
    thanks for the help. i actually realized what was going on while on the way to the grocery store. p = F(net) dt -- if f is changing every t the area beneath the curve at that point would be the net force acting at t -- therefor by integrating from 0 to 2 i'm getting the total amount of force from that duration. from the net force i can determine velocity based on the change in momentum.

    duh, i'm retarded

    thanks again for the help.
     
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