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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
    vo=8.50m/s
    k=68.00N/m
    μ=.200
    x=?


    2. Relevant equations
    Kf-Ki=W
    K=1/2mv2
    Fs=-kx
    Ffric=μNf
    W=Fd



    3. The attempt at a solution
    First I found the work of the whole system (friction and spring forces) by using
    Kf-Ki=W.
    For the initial kinetic energy, I used
    K=1/2mv2=155J.
    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
    W=-155J
    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:
    W=(-kx+μNf)x
    I work it out to a quadratic by plugging the numbers in...
    -68x2+8.44366x+155=0
    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

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    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

    tiny-tim

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    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

    tiny-tim

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    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!!!
     
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