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Homework Help: What is the final velocity of the particle?

  1. Sep 28, 2011 #1
    The image below "gives, as a function of time t, the force component x that acts on a 2.40 kg ice block that can move only along the x axis. At t = 0, the block is moving in the positive direction of the axis, with a speed of 2.5 m/s. What is its velocity, including sign, at t = 11 s?


    Answer: 8.8 m/s"

    Relevant equations:
    Vf = Vo + at

    I was thinking that the total area under the graph would give the net force the component seeing that the y-axis represents Force and the x-axis represents Time. With that I got the net force 15N (29N - 14N). So,

    F = 25N
    F = ma

    m = 2.40kg
    Vo = 2.5m/s
    t = 11
    Vf = ?

    Therefore, with the given information, I decided to use the following kinematic equation to find acceleration.
    Vf = Vo +at
    a = (Vf - Vo)/t

    F = m(Vf - Vo)/t

    And if we solve for Vf we get
    Vf = tF/m + Vo

    However, when I plug in the units I get an answer that is way way off.
    Vf = 11(15)/2.4 + 2.5
    Vf = 165/2.4 + 2.5
    Vf = 68.75 + 2.5
    Vf = 71.25 m/s

    I know I added in the units at the end but I checked and the units come out correctly. Therefore, my question is: what did I do wrong? Why am I so far off? Obviously I did something horribly wrong in my calculation, but where?

    Thank you for taking the time review my question. Any help is much appreciated.
  2. jcsd
  3. Sep 28, 2011 #2


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    Homework Helper

    This is a Force time graph. If you factor in the mass, you have an acceleration time graph. The area "under" and acceleration time graph gives change in velocity does it not?
    Don't forget it had an initial velocity.

    EDIT: Alternately, the area under a Force-time graph gives you the impulse of the force. The impulse gives you the change in momentum, which can easily be changed to a change in velocity.
    Last edited: Sep 28, 2011
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