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Conservation of Energy

  1. Sep 21, 2007 #1
    [SOLVED] Conservation of Energy

    I know, this problem is making me feel dumb but I can't figure it out:

    A crossbow takes 50.0 pounds of force to draw the arrow back by 12.5 inches, and the weight of the arrow is 2 ounces. What is the speed of the arrow when it is released?
    (1 lb = 4.448 N; 1 in = 2.54 cm; 1 oz = 28.35 grams)

    Could someone show me how to solve this?? Thanks to whoever helps.

    *It is not momentum, but conservation of energy, sorry.
    Last edited: Sep 21, 2007
  2. jcsd
  3. Sep 21, 2007 #2


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    This is a homework problem in Introductory Physics. And please do not do multi-posting.

    With the force and displacement, one can determine a spring constant. The force usually decreases with displacement, unless the bow is specially designed for constant force.

    Knowing the force one can determine the acceleration of the mass (arrow), and from the acceleration the velocity.

    The other part as mgb_phys pointed out is that the energy stored in the bow (spring) must equal the kinetic energy of the mass when the bow string reaches its rest (zero) position.

    And the work released = F*d, where F is the force and d is the displacement. But the question then becomes, is F a function of displacement (x), where 0 < x < d.
    Last edited: Sep 21, 2007
  4. Sep 21, 2007 #3


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    How much energy does the crossbow have before the arrow is released? ie: how much work has been done on the crossbow?
  5. Sep 21, 2007 #4


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    Work is force * distance and kinetic energy is 1/2mv^2.
    So assuming all the energy goes into the arrow it's pretty simple
  6. Sep 21, 2007 #5


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    Why have you posted this exact question 4 times on the forums?
  7. Sep 21, 2007 #6

    Doc Al

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    Threads merged--DO NOT multi-post!
  8. Sep 21, 2007 #7
    Still not getting it for some reason...

    I found the spring constant (k) to be -700.472

    I found the acceleration (a) to be 392.24 m/s/s

    I found the total work (W) on the system before the arrow is shot to be 70.612 J
    W = Fd = 50(4.448)(.3175m) = 70.612 J

    I am not sure what to do from here, the only thing that I do know is:

    (1/2)mv^2 + mgh + (1/2)kx^2 = (1/2)mv^2 + mgh + (1/2)kx^2

    Do I need to even use this equation? Setting Work equal to (1/2)mv^2 does not give me the correct answer. What am I not doing correctly? Is there energy that I am missing.

    I know this is a simple problem and I have NO IDEA why it is causing me so much trouble. Will someone please help so more?
  9. Sep 21, 2007 #8


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    (1/2)kx^2 = 32.538J

    set (1/2)mv^2 = 32.538

    m = 0.0567kg

    v= 33.88m/s ? I think this is the way to do it.

    If 50lb, is the average force in pulling the arrow back then...

    70.612J = (1/2)mv^2

    v = 49.9m/s. Is that what you got?

    EDIT: oops... should be 35.29m/s as ryo calculated. I used 12inches, instead of 12.5inches.
    Last edited: Sep 21, 2007
  10. Sep 21, 2007 #9

    Instead of using .0567 m, I was using .567 m. Oh well, I stressed over such a small mistake. So, I got

    (1/2)kx^2 = (1/2)mv^2

    (1/2)700.472(.3175)^2 = (1/2)(.0567)v^2

    v = 35.29 m/s !!!!!!

    Thanks for the input guys, it all made sense. I just incorrectly converted g into kilograms. Maybe I was thinking in terms of m to cm and dividing by 100 instead of 1000. Thanks though guys!
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