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Math Help - arrow problem, flying horizontally

  1. #1
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    arrow problem, flying horizontally

    A 0.062 kg arrow is fired horizontally. The bowstring exerts an average force of 65 N on the arrow over a distance of 0.75 m. With what speed does the arrow leave the bow?

    need in m/s

    is this just a simple equation problem, to find velocity?
    or F = mass * acceleration?
    or just need to find velocity?

    the part that says horizontally, is confusing me.
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  2. #2
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    Would it be faster or slower if you fired it down or up? The acceleration due to gravity is perpendicular to the bow acceleration when fired horizontally, no?
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  3. #3
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    Thanks, i just used F= ma to find the acceleration where N = 65 and .062 kg was the mass.

    then i used the kinematics equation of v^2 = v0^2 + 2*a * (xf - x0)

    and plugged everything in to get the velocity after square rooting the sides.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by rcmango View Post
    A 0.062 kg arrow is fired horizontally. The bowstring exerts an average force of 65 N on the arrow over a distance of 0.75 m. With what speed does the arrow leave the bow?

    need in m/s

    is this just a simple equation problem, to find velocity?
    or F = mass * acceleration?
    or just need to find velocity?

    the part that says horizontally, is confusing me.
    Quote Originally Posted by rcmango View Post
    Thanks, i just used F= ma to find the acceleration where N = 65 and .062 kg was the mass.

    then i used the kinematics equation of v^2 = v0^2 + 2*a * (xf - x0)

    and plugged everything in to get the velocity after square rooting the sides.
    You can do it this way, though I suspect you are now learning about the Work-Energy Theorem.
    W = \Delta KE

    The work done is simply W = Fs since the force and displacement are in the same direction. \Delta KE = KE - KE_0 = \frac{1}{2}mv^2 since the initial speed of the arrow is 0 m/s. Thus
    Fs = \frac{1}{2}mv^2

    Note: Here I am using a constant F. In simple cases we can get away with using an average force as a constant force over a time interval. (Perhaps in all cases. I'm trying to be conservative as I'm too tired to think through details right now.)

    -Dan
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    not sure i'm following correctly, but when i used your formula for F = 1/2 * m * v^2

    i get something around 45 m/s.

    which is not correct. not sure what i've done there.

    however, 39.6 m/s is correct.
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by topsquark View Post
    Fs = \frac{1}{2}mv^2
    Quote Originally Posted by rcmango View Post
    not sure i'm following correctly, but when i used your formula for F = 1/2 * m * v^2

    i get something around 45 m/s.

    which is not correct. not sure what i've done there.

    however, 39.6 m/s is correct.
    You missed the "s" which is 0.75 m. (Check the units on both sides of the equation you posted. You will see that this cannot be a correct equation.)

    -Dan
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