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Conservation of Energy with Friction

  1. Mar 28, 2009 #1
    1. The problem statement, all variables and given/known data
    A 179 g block is launched by compressing a spring of constant k=200 N/m a distance of 15 cm. The spring is mounted horizontally, and the surface directly under it is frictionless. But beyond the equilibrium position of the spring end, the surface has coefficient of friction [tex]\mu[/tex] = 0.27. This frictional surface extends 85 cm, followed by a frictionless curved rise, as shown in the figure:

    Code (Text):
                            |
                           /
    |xxxx|[ ]             /
    ---------=========---/
             ^           ^---- Imagine this as a bottom right quarter of a circle.
             |                     Things go up and slide down with no friction
             |-- Friction
                   85 cm
    After launch, where does the block finally come to rest? Measure from the left end of the frictional zone.

    2. Relevant equations

    Conservation of Energy:
    [tex]KE_i + PE_i + WE_i= KE_f + PE_f + WE_f[/tex]

    3. The attempt at a solution

    So, values:

    m = 0.179 kg

    k = 200 N/m
    x = -0.15 m

    [tex]\mu[/tex] = 0.27
    [tex]d_{friction}[/tex] = 0.85 m
    [tex]F_{normal}[/tex] = mg = 0.179 * 9.8

    ----

    The initial total energy is the spring's potential energy. All other initial energies are zero. So:

    [tex]PE_i = 1/2kx^2 = 1/2(200)(0.15) = 2.25 J[/tex]

    Then, I proceeded to calculate the energy loss due to friction, by the amount of work done:

    [tex]WE_{per slide} = F_{friction} d_{friction} = \mu F_{normal} d_{friction} = 0.27 \times (0.179)(9.8) \times 0.85 = 0.40258 J[/tex]

    I found out that, if I divide the first by the second, it takes the 6th slide (back and forth) to actually "stop" the box from sliding, but at this point I'm really lost. The remaining energy might not be enough, for example, to overcome the static friction, and I'm unsure as to how to approach this problem anymore.

    Any help would be appreciated. This problem is a bonus problem in one of my assessments, and I don't really need it solved to pass. But I'd really love to know how to solve it.
     
    Last edited: Mar 28, 2009
  2. jcsd
  3. Mar 28, 2009 #2
    Actually, never mind me. I solved it with a friend's help. Yay!

    If you divide the first by the second, the remainder is 0.237 J. Using that information, I was able to calculate the distance travelled, 50cm. So, subtract that from 85cm, it stopped at 35cm from the left of the frictional surface.

    (Work = Force * distance, or 0.237 = Friction Force * distance.)
     
  4. Mar 28, 2009 #3

    Doc Al

    User Avatar

    Staff: Mentor

    Good.

    You found that it makes 5 complete trips across the friction patch and one partial trip. Find the length of that partial trip.
     
  5. Mar 28, 2009 #4
    Thanks! That's exactly what I had to do.

    As a side question, though, how does this problem relate to the concept of static friction and dynamic friction?
     
  6. Mar 28, 2009 #5

    Doc Al

    User Avatar

    Staff: Mentor

    Static friction doesn't play a role in this problem.
     
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