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Potential Energy (WebAssign Question)

  1. Mar 16, 2009 #1
    1. The problem statement, all variables and given/known data

    A 4.8 g marble is fired vertically upward using a spring gun. The spring must be compressed 8.0 cm if the marble is to just reach a target 27 m above the marble's position on the compressed spring.

    (a) What is the change in the gravitational potential energy of the marble-Earth system during the 27 m ascent?
    (a) ? J

    (b) What is the change in the elastic potential energy of the spring during its launch of the marble?
    (b) ? J

    (c) What is the spring constant of the spring?
    (c) ? N/m

    2. The attempt at a solution

    (a)
    27 - .080 = 26.92
    (26.92)(9.8)(.0048) = (1/2)(.0048)v^2
    v = 22.97 <<<< Isn't this the answer? But it's wrong.

    (b)
    (1/2)(the answer to c: ?)(.080)^2

    (c)
    (1/2)(.0048)(the answer to a: 22.97?)^2 + (.0048)(9.8)(.080) = (1/2)(k)(.080)^2
    k = ?
     
  2. jcsd
  3. Mar 16, 2009 #2

    rl.bhat

    User Avatar
    Homework Helper

    a) mgh
    b) 1/2*k*x^2
    From a) and b) you can calculate c)
     
  4. Mar 16, 2009 #3
    What do you state gravity as?

    the gravitational potential energy equation is this:

    [tex]U_g=mgh[/tex]

    "U" is the energy in joules, "m" is mass in kg, "g" is the gravity value in ms-2, and "h" is the maximum height of the object in metres. the value of "g" is normally taken as 9.8 or 10ms-12.

    That might clear some of the errors up. :)
     
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