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Sprin compression

  1. Aug 8, 2007 #1
    1. The problem statement, all variables and given/known data
    Could somebody help walk me through this problem?:
    A spring has its right end fixed and is installed on a horizontal table so that the free end, in equilibrium, is at x= 3m. A 1.65 kg block coming from the left slides along the table. When it passes the origin, it is moving at 5.58 m/s. It strikes the spring, compresses it momentarily, and is then sent back toward the left, where it eventually comes to rest at the point x= 1.5m. The coefficient of kinetic friction betweeen the block and the table is .300. By what distance was the spring compressed?

    2. Relevant equations
    F(of kinetiic friction)=(coefficeint of kinetic friction)(m)(g)
    KE=.5(mass)(velocity squared)

    3. The attempt at a solution
    KE=.5(1.65kg)(5.58 m/s squared)
    F(of kinetic friction)=.3(1.65)(9.8)
    =4.851 N
  2. jcsd
  3. Aug 8, 2007 #2


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    Homework Helper

    You seem to have ended up with a force not a distance!

    You had almost the correct approach, energy is conserved
    First it is the KE of the incoming block.
    Then some is lost in force*distance of the friction on the table.
    Then it is the force*distance of the compressed spring (when the block is at rest) then it is the force*distance the retreating block travels.
  4. Aug 8, 2007 #3


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    I'd use this idea... Work done by friction = final energy - initial energy

    The distance travelled into the spring is a variable in the left side of the equation.
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