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A spring and its coefficient

  1. Oct 2, 2008 #1
    1. The problem statement, all variables and given/known data
    A block of mass 1 kg is forced against a horizontal spring of negligible mass, compressing the spring an amount x1= 0.2m. When released, the block moves on a horizontal tabletop a distance x2 = 1.0m before coming to rest. The spring constant k is 100 N. What is the coefficient of friction, μ, between the block and the table?

    2. Relevant equations
    K1 + U1 = K2 + U2
    fs= -kx
    Kinetic energy = 1/2kv2
    Potential energy = mgh

    3. The attempt at a solution
    At x1 kinetic energy is 0 and potential energy is 1/2kΔx2
    So the total energy would be = 40.
    However, I do not know how to tie this in with the coefficients of friction.
  2. jcsd
  3. Oct 3, 2008 #2
    potential energy of a linear spring = [tex] \frac{1}{2} kx^2 [/tex]

    therefore potential energy of this spring at this compression = [tex] \frac{1}{2} (100) (.04) = 2J [/tex]

    conservation of momentum [tex] E_k = E_p [/tex]

    so kinetic energy also = 2J at the moment that the spring reaches full length. [tex] \frac{1}{2}mv^2 = 2 [/tex]

    Now solve for velocity at the instant it leaves the spring. Calculate the force due to gravity on the block.
    Let me know how you go from there, it should be clear how to finish the question.
    Last edited: Oct 3, 2008
  4. Oct 3, 2008 #3
    Okay I understand now, thank you!
  5. Oct 3, 2008 #4
    No worries, I just realised that there was an error in the equation I wrote out on the second line of my post, however the result is still the same. I have corrected it now. apologies.
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