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Problem involving Newton's 2nd law of motion

  1. Sep 24, 2007 #1
    1. The problem statement, all variables and given/known data

    An arrow, starting from rest, leaves the bow with a speed of 25.0m/s. If the average force exerted on the arrow by the bow were doubled, all else remaining the same, with what speed would the arrow leave the bow?



    2. Relevant equations

    F = ma


    3. The attempt at a solution

    For a force, F, the launch velocity of the arrow was 25.0m/s. Acceleration is F/m and multiplying F/m by a time t gives velocity at time t. So I re-wrote the first scenario as 25.0m/s = t/m * F. In the second scenario the force is doubled (2F), so I re-wrote the velocity as Xm/s = t/m * 2F. Since t and m remain the same for both scenarios, is the velocity of the 2nd scenario simply that of the 1st scenario's but doubled (50m/s)?
     
  2. jcsd
  3. Sep 24, 2007 #2

    Astronuc

    User Avatar

    Staff: Mentor

    One has to ask, what does that mean? i.e. the time remains the same or the displacement of the draw.

    Doubling the force on the same mass would double the acceleration, and for the same the displacement, that would reduce the time of the launch.
     
  4. Sep 27, 2007 #3
    more detail

    By all else remaining constant, it means that the arrow's displacement (x) remains the same. That seemed to be the general class consensus. The acceleration doubles when the force is doubled, that makes sense, but the next step (actually finding the velocity of such an arrow) is giving me some trouble.

    What I've done is try to solve some of the kinematic equations to get expressions for displacement and time from which I can calculate velocity; I am running into troubles, because for the first arrow (acceleration = a) I can't get the velocity to come out as 25m/s, i always get half that.

    This is the approach I am taking:

    First arrow: V0 = 0m/s, V = 25m/s, Acceleration = a, Displacement = x

    Finding x in terms of a:

    V^2 = V0^2 + 2ax
    625 = 2ax
    312.5/a = x

    Finding t in terms of a:

    V = V0 + at
    25 = at
    25/a = t

    Finding 1st a using new expression for and t:

    V = V0 + at
    25 = 312.5/a * 25/a
    25 = 7812.5/a^2
    a^2 = 7812.5/25
    a = sqrt(7812.5/25) = 17.68 m/s2

    check: V = 17.68 * 25/17.68 = 25m/s

    Using initial acceleration*2 to find velocity after a doubled force (acceleration)

    V = V0 + at
    V = (17.68*2) * 25/(17.68*2)
    V = 35.36 * 25/35.36
    V = 25

    The same velocity? The expressions seem to work ok for the 1st arrow's conditions, 17.68m/s acceleration matches the 25m/s launch velocity given in the problem. Can anyone help me figure out what's going wrong?
     
  5. Sep 27, 2007 #4

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    Why did you substitute the expression for x as the value for a? If you had kept the units you would have seen that what you did here is invalid. So what are the units?

    Good so far; you even have units on the initial velocity (0 m/s rather than 0).

    Now you are dropping units. This would be much better done as

    [tex]\begin{aligned*}
    v^2 &= v_0^2 + 2 a x \\
    2 a x &= 625\; \text{m}^2/\text{s}^2 \\
    x &= \frac{312.5\; \text{m}^2/\text{s}^2}{a} \\ & \\
    v &= v_0 + at \\
    at &= 25 \text{m/s} \\
    t &= \frac{25\; \text{m/s}}{a}
    \end{aligned*}[/tex]

    There is no way to find a numerical solution for the acceleration a because you don't know the displacement x. You can, however, still determine what would happens when the acceleration doubles.
     
    Last edited: Sep 27, 2007
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