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Homework Help: Spring question(just need answer checked)

  1. Mar 25, 2010 #1
    1. The problem statement, all variables and given/known data

    (a)A spring is attached to a ceiling, and has a relaxed length of 25 cm. When a mass m = 0.80 kg is attached to the spring it stretches to an equilibrium length of L0 = 34 cm.

    (b) I lift the mass until the spring returns to its relaxed length, and then release it. When the mass returns to the equilibrium length, what is its speed?
    (c) After I release the mass and it falls, what is the length of the spring when the mass reaches its lowest point?

    2. Relevant equations

    .5mv^2=kinetic energy
    mgy=potential energy

    3. The attempt at a solution
    so K=87.2 for part A

    Sorry for the missleading title but i realized im not sure how to do part b or c, i believe for part b i have to set mgy+.5mv^2=mgy+.5mv^2 but im not entirely sure

    Want to make sure im doing this right before my quiz tommorow, thanks.
    Last edited: Mar 25, 2010
  2. jcsd
  3. Mar 27, 2010 #2


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    Homework Helper

    Your calculation of part A is correct.
    Consider the equilibrium position as the reference point for measuring PE with zero initial PE.
    When you lift the mass through a distance x, the rise in PE = mgx.
    When you release the mass its initial velocity is zero.
    When it crosses the equilibrium position, fall in PE = mgx
    This change in PE in converted in to KE of mass and elastic potential energy in spring.
    Write down the equation and solve for v.
    For the last part, the final velocity of the mass is zero.
    So change in KE = fall in PE + increase in the elastic PE
    Solve this equation to find the final extension of the mass..
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