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Essay question: motion

  1. Nov 20, 2007 #1
    An arrow, starting from rest, leaves the bow with a speed of 25.0 m/s . If the average force exerted on the arrow by the bow were double, all else remaining the same,with what speed would the arrow leave the bow
     
  2. jcsd
  3. Nov 20, 2007 #2
    First case -

    Fd = 1/2mv^2
    v^2 = 2Fd/mv
    v = sqrt(2Fd/mv)
    25 = sqrt(2Fd/mv) - Equation (1)

    Second case
    (2F)d = 1/2mv^2
    v^2 = 4Fd/mv
    v = sqrt(4Fd/mv)
    = sqrt(2)*sqrt(2Fd/mv)
    = sqrt(2) * 25 (from (1))
    v = 35.4 ms-1(3sf)

    I think :D
     
  4. Nov 20, 2007 #3
    Fd = 1/2mv^2?
     
  5. Nov 20, 2007 #4
    isnt it d=1/2av^2
    d=1/2(f/m)v^2
    md=1/2(f)v^2?
     
  6. Nov 20, 2007 #5
    Hmm im not familiar with those equations...

    Work done is Force x Distance.

    Since all the work goes into the kinetic energy of the arrow -

    Work Done = Kinetic Energy
    Fd = 1/2*mv^2

    Yeah?
     
  7. Nov 20, 2007 #6
    hey, i never learned work done equations, we only used the big five kinematic equations
     
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