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Average Forces and Newtons Laws

  1. Sep 28, 2006 #1
    I have 2 questions that I am approaching, one seems very simple but I am still at a loss. The other I have attempted multiple times to no avail. Here they are:

    1. An arrow, starting from rest, leaves the bow with a speed of 28.9 m/s. If the average force exerted on the arrow by the bow was increased 4 times and the arrow was accelerated over the same distance, then with what speed would the arrow leave the bow?

    I know that the Sum(F) = ma, but since we only have the Vo here I do not know where to start, I have tried multiplying my figures by 4, but that does not work either.

    2. Airplane flight recorders must be able to survive catastrophic crashes. Therefore, they are typically encased in crash-resistant steel or titanium boxes that are subjected to rigorous testing. One of the tests is an impact shock test, in which the box must survive being thrown at high speeds against a barrier. A 50-kg box is thrown at a speed of 320 m/s and is brought to a halt in a collision that lasts for a time of 7.1 ms. What is the magnitude of the average net force that acts on the box during the collision?

    My work: What I did here was use the equations of Kinematics, but I am not sure if that is correct.

    What we know:

    m = 50kg
    Vo = 320 m/s
    V = 0
    t = 7.1 (milliseconds) ms

    I used the equation:

    V=Vo + a(.71) <------ I converted that to seconds

    0 = 320 + a(.71) = a = 450 m/s^2

    I used a in this equation

    a = F/m, so F = ma, which was 50kg(450 m/s^2) = 22500 N

    This was wrong.

    Any help is appreciated, thanks.
  2. jcsd
  3. Sep 28, 2006 #2

    Andrew Mason

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    This is an energy question. Are you familiar yet with energy? If so, what is the Kinetic Energy of the arrow? How is that related to the force applied to the arrow and the distance over which it is applied?

    If you have not taken energy yet, you can still solve this but it is a little harder. What is the speed of the arrow in terms of the force and time (how is momentum related to impulse)? Since [itex]d = v_{avg}t = \frac{1}{2}v_{f}t[/itex] we have: [itex]t = 2d/v_f[/itex]. Substitute that expression for t in the momentum/impulse expression to express v in terms of the force and the distance through which the force acts.

    Your method is right but you are using 7.1 seconds instead of 7.1 milliseconds. So you are out by a factor of 1000.

  4. Sep 28, 2006 #3

    Thank you very much, I have not taken energy yet so I don't know how to do it that way, plus your explanation still confused me :biggrin:

    but nonetheless, thanks so much for the time, I am going to try the second question now.
  5. Sep 28, 2006 #4

    Andrew Mason

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    I was trying to get you to realize that the change in momentum of an object is the force x time over which the force is applied:

    [tex]m\Delta v = f_{avg}\Delta t[/tex]

    This is simply a consequence of f=ma:

    [tex]f_{avg} = ma_{avg} = m\Delta v/\Delta t[/tex]

    So if you substitute [itex]\Delta t = 2d/v_f[/itex], you get:

    [tex]m\Delta v = mv_f = f_{avg}\Delta t \rightarrow \frac{1}{2}mv_f^2 = fd[/tex]

    So if you multiply the force by 4 what happens to the velocity?

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