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Homework Help: Why can't i do it my way

  1. May 6, 2010 #1
    A 400 g particle moving along the x-axis experiences the force shown in Figure Ex11.15. The particle goes from vx = 2.5 m/s at x = 0 m to vx = 5.5 m/s at x = 2 m. What is Fmax?

    my solution manual told me to do this with work and energy

    but why can't i do
    kinematic:
    5.5^2=2.5^2+2A2=6

    F=ma
    (0.4KG)(6m/s^2)=.5Fmax(2)=2.4N
    area under the curve--triangle---

    but the correct answer is 4.8N.........ugh
     

    Attached Files:

  2. jcsd
  3. May 6, 2010 #2
    Wait the area underneath the curve is work right and my answer is the total force? ..... so i have to do W=(F)(delta X)???????
     
  4. May 6, 2010 #3
    If the curve had the force as a function of time then your approach would work.

    [tex] m(v_2-v_1) = \int_{t_1}^{t_2} F(t) dt [/tex]

    this is simply F=ma integrated

    but since it isn't you can't do this and you will have to use

    [tex] W = \int_{x_1}^{x_2} F(x) \cdot dx [/tex]
     
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