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Find displacement of marble launched from a spring (Energy), help please

  1. Oct 30, 2008 #1
    1. A horizontal spring, of constant 12N/m, is mounted at the edge of a lab bench to shoot marbles at targets on the floor 93.0 cm below. A marble of mass 8.3 x 10^-3 kg is shot from the spring, which is initially compressed a distance of 4.0 cm. How far does the marble travel horizontally before hitting the floor?

    2. Relevant equations
    E = 0.5kx^2
    E = 0.5mv^2
    delta d = 0.5(acceleration)(delta time)^2
    dx = vx x t

    3. The attempt at a solution

    E = 0.5kx^2
    = 0.5(12N/m)(0.04m)^2
    = 0.0096J

    E = 0.5mv^2
    0.0096J = 0.5(8.3x10^-3kg)(v)^2
    Square root of [0.0096J / (0.5x8.3x10^-3kg)] = v
    2.31 m/s = v

    delta d = 0.5at^2
    t = square root of [ (2 times delta d) / a] *accleration = gravity, 9.8m/s
    t = square root of [ (2 x 0.93m) / (9.8m/s^2)]
    t = 0.436s

    dx = vx x t
    = 2.31 m/s x 0.436s
    = 1.00716m

    I've got 1.00716m as the answer. However, the textbook says the answer is 0.66m. I'm guessing I did something wrong at finding the velocity. I can't seem to figure this problem out. Please help. Thanks in advance! :smile:
  2. jcsd
  3. Oct 30, 2008 #2


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    Homework Helper

    Welcome to PF.

    2.31 is V2 not V
  4. Oct 30, 2008 #3
    Thank you for the welcoming! :biggrin:

    And yes, I am a clutz, for forgetting to square root... :blushing::redface: Gosh this is embarrasing...

    Thanks so much for the help!
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