1. Not finding help here? Sign up for a free 30min 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!

Find Spring constant given distance stretched by placing rock on spring

  1. Nov 18, 2012 #1
    1. The problem statement, all variables and given/known data
    A 8.00 kg stone lies at rest on a spring. The spring is compressed 10.0 cm (.1 m) by the stone. What is the spring constant?

    2. Relevant equations
    F = -kx
    E = (1/2)kx^2
    U = mgh

    3. The attempt at a solution
    The solution provided uses a net force equation balancing Hooke's Law with mg:
    -kx = mg
    k = mg/x = 784 N/m

    When I first tried this problem, I approached it using conservation of energy:
    U0 + E0 = Uf + Ef
    Ef = U0 + E0 - Uf = U0 + 0 - 0 = U0 = 7.84 J
    Where t=0 is the rock at the top of the uncompressed spring and t=f is the rock in equillibrium on the compressed spring.
    k = 2Ef/x^2 = 1568 N/m

    Using my approach, I am off by a factor of 2. Where is my misunderstanding here? I understand the solution involving net force, but shouldn't the conservation of energy work out too?
     
  2. jcsd
  3. Nov 19, 2012 #2
    The weight of the stone is balanced by the restoring force of the spring when it is compressed all the way.
     
  4. Nov 19, 2012 #3

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    When you used the energy approach with no kinetic energy in the initial anf final position and just the conservative weight and spring forces acting, you assumed that the rock atop the spring was released from rest. When you do it this way, the spring stretches beyond the equilibrium and to a maximum deflection until it momentarily stops, at which point it starts back up and continues to oscillate back and above and below the equilibrium position. This gives you an incorrect result because you assumed that the compression of 0.1 m occurs at the bottom of the motion, not at the equilibrium position. For the problem at hand, the rock is not released, rather, it is slowly lowered by an external variable force applied by your hand up to the equilibrium position where you then let go. You did not account for this force or the work it does in your equation.
     
  5. Nov 19, 2012 #4
    Thank you PhanthomJay. Clear as daylight.
     
  6. Nov 20, 2012 #5
    One can just use F=kx in this case.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook