Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Distance spring decompresses when friction is involved

  1. Feb 9, 2012 #1
    1. The problem statement, all variables and given/known data
    A block of mass m = 4.30 kg slides along a horizontal table with speed v0 = 8.50 m/s. At x = 0 it hits a spring with spring constant k = 68.00 N/m and it also begins to experience a friction force. The coefficient of friction is given by μ = 0.200. How far did the spring compress when the block first momentarily comes to rest?
    m=4.30 kg

    2. Relevant equations

    3. The attempt at a solution
    First I found the work of the whole system (friction and spring forces) by using
    For the initial kinetic energy, I used
    For the final kinetic energy, I used 0, because I'm measuring the distance after the mass came to a stop. So, plugging the numbers into the Work equation, I simply get
    I'm pretty sure I'm on the right track so far?
    Then, using
    W=Fd (just multiplying the forcexdistance because both vectors are in the same direction)
    I plugged in:
    I work it out to a quadratic by plugging the numbers in...
    I solve then for x (using the quadratic root finder on my calculator, so I'm not screwing up the quadratic..), but both of the answers it gives me are incorrect..
    What am I doing wrong? Help please!!
  2. jcsd
  3. Feb 9, 2012 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Careful: That only works when the force is constant. The spring force is not constant.

    Hint: What's the energy stored in a compressed spring?

    Also: Careful with signs. The work done on the mass by both the spring and the friction force is negative.
  4. Feb 9, 2012 #3


    User Avatar
    Science Advisor
    Homework Helper

    welcome to pf!

    hi becky! welcome to pf! :smile:

    isn't it 1/2 kx2 ? :wink:
  5. Feb 9, 2012 #4
    Re: welcome to pf!

    W = ∫Fdx (F dot dx, technically)
    Fspring = kx

    W = ∫kxdx
    Wspring = kx2/2

    Just to show where it came from, since W does NOT equal Fx, that is not a definition, it is a derivation of work for a constant force.
    Last edited: Feb 9, 2012
  6. Feb 12, 2012 #5
    Thank you guys! I finally figured out the right answer! But when I solve for the quadratic equation, why is the distance vector only distributed to the friction force? This is what I tried to do...
    W=(-1/2kx^2+friction force)*x
    and this is what is right...
    W=-1/2kx^2+friction force*x
    Is it because the work of the spring is already in terms of "work" and the friction force isn't?
  7. Feb 12, 2012 #6


    User Avatar
    Science Advisor
    Homework Helper

    hi becky! :smile:
    yes!! :smile:

    work done is force "dot" distance, and 1/2 kx2 is 1/2 kx (average force) times x (distance)! :wink:
  8. Feb 13, 2012 #7
    Thank you!!!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook