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Arrow is short vetically upwards (Force/Springs)

  1. Mar 29, 2009 #1
    1. The problem statement, all variables and given/known data

    A 110 g arrow is shot vertically from a bow whose effective spring constant is 500 N/m

    If the bow is drawn 55 cm before shooting the arrow, to what height does the arrow rise?


    2. Relevant equations

    F=ma
    U=kx^2 - for Spring (bow)
    F=-kx

    KE=1/2 mv^2
    W=fs

    3. The attempt at a solution

    I found by using:
    F=500 * .55
    = 275N

    The arrow moves upwards at 275N

    It therefore accelerates at:
    F=ma
    a= F/m
    = 275/.110

    Gravity is assumed to be -9.8.

    Thanks for any help
     
    Last edited: Mar 29, 2009
  2. jcsd
  3. Mar 29, 2009 #2

    Pythagorean

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    Gold Member

    That arrow does not move upwards with 275 newtons. The bow puts 275 Newtons of upward force on it...

    So what is the total acceleration on the arrow after you factor in gravity and compute the acceleration from the spring (275/.110)?
     
  4. Mar 29, 2009 #3
    I figured that much out but could you explain further. Is it simply like vectors? Where I just minus 9.8 from 2500 and then use 2491.2 as the overall acceleration upwards?
     
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