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Elastic potential energy - springs

  1. Oct 29, 2012 #1
    I know this problem has been asked before but i am trying to understand.

    1. The problem statement, all variables and given/known data

    A 2.00-kg block is pushed against a spring with negligible mass and force constant k = 400 N/m, compressing it 0.220 m. When the block is released, it moves along a frictionless, horizontal surface and then up a frictionless incline with slope 37 degrees

    (a) What is the speed of the block as it slides along the horizontal surface after having left the spring?

    2. Relevant equations

    K1 + U2 = K2 + U2
    Vel = 1/2 kx^2

    K = 1/2mv^2

    3. The attempt at a solution

    If i think of K1 and U1 as the spring compressed then K2 + U2 is the box leaving the spring.
    If that logic is correct.?

    My teacher has taught us to do this:

    K1 =
    U1 =
    K2 =
    U2 =

    Now, i am confused on how to fill it in. Then, i can use K1 + U2 = K2 + U2.
    can you help me fill in and explain why? I think this will help me understand how to do the problem.

    In my mind potential energy is converted to kinetic energy? So, saying K1 + U2 is spring compressed
    K1 = 0
    U1 = 1/2 kx^2
    K2 = 1/2 mv^2
    U2 = 0
    Last edited: Oct 29, 2012
  2. jcsd
  3. Oct 29, 2012 #2
    You've done it right.
    The box isn't moving initially, so the kinetic energy is 0.
    The box isn't compressing the spring when it leaves, so the potential energy is 0.
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