1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Elastic potential energy - springs
Loading...