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Springs and kinetic energy

  1. Jun 5, 2010 #1
    1. The problem statement, all variables and given/known data

    A relaxed spring with spring constant k = 70 N/m is stretched a distance di = 63 cm and held there. A block of mass M = 7 kg is attached to the spring. The spring is then released from rest and contracts, dragging the block across a rough horizontal floor until it stops without passing through the relaxed position, at which point the spring is stretched by an amount df = di/7.

    2. Relevant equations

    W = F*d
    KE = 1/2 k*x^2
    KE = 1/2 m*v^2

    3. The attempt at a solution

    All I need is to find the change in kinetic energy, but i havent been able to get the right value yet.

    I tried to find the change in kinetic energy by using 1/2kx^2, using .63 m as x, also trying 0.63 - 0.09 m as x, but i havent gotten the right value for change in KE.

    Please help.
     
  2. jcsd
  3. Jun 5, 2010 #2

    rock.freak667

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    What exactly do you need to find? Because they said the surface was 'rough' means that energy is lost due to friction.
     
  4. Jun 5, 2010 #3
    Well, in order to answer this I need to find the Work done by friction, which means I would first need the Work done by the spring, but I'm not sure how to find that.
     
  5. Jun 5, 2010 #4

    rock.freak667

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    Right then.

    Well you do want to find a change in energy, but it is not kinetic energy.

    Initially it is stretched, so the mass will have 0 KE. At the final value of compression (di/7), the mass also has 0 KE. But there is a change in energy, which you calculated. Shouldn't this change in potential energy be associated with the frictional force?
     
  6. Jun 5, 2010 #5
    Yes, but I have not idea how to find that change in energy.

    I tried deltaE = 1/2kx^2 (initial) - 1/2kx^2 (final)

    If that's not right then I have no idea.

    But then wouldn't that change in energy = Work by friction?
     
  7. Jun 5, 2010 #6

    rock.freak667

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    Yes but what are you using as xfinal and xinitial?
     
  8. Jun 5, 2010 #7
    Efinal = 1/2*(70)*(.63/7)^2
    Einitial = 1/2*(70)*(.63)^2

    xfinal = .63/7 m
    xinitial = .63 m

    would the equation then be, Work by friction = M*(Normal)*(.63/7) = Efinal - Einitial ?
     
  9. Jun 5, 2010 #8
    The work done by friction would be equal to the change in the Spring's potential energy. In a frictionless situation, this drop in potential energy would translate into an increase in kinetic energy. Since we have friction, the block is slowed down and the energy that initially went to kinetic energy is eventually taken away by friction.
     
  10. Jun 5, 2010 #9
    Then how do you find the change in potential energy? The only equation I know is PE = mgh, and since the h does not change this is no help.
     
  11. Jun 5, 2010 #10

    rock.freak667

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    Yes, the work done by friction should be that, unless I am missing something. Although the distance the block travels is no 0.63/7 m
     
  12. Jun 5, 2010 #11
    no, but that is the final distance. the total distance traveled would be .63 - .64/7, right?

    does that affect the equation?
     
  13. Jun 5, 2010 #12
    Note that the potential energy is actually

    [tex]U_{spring}(x)=\frac{1}{2}k(x-x_{equilibrium})^{2}[/tex]

    but we can set [tex]x_{equilibrium}=0[/tex] in this case so that what you have above is ok.
     
  14. Jun 5, 2010 #13

    rock.freak667

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    It does not affect the change in potential energy part.

    From equilibrium to initial is 0.63m. From equilibrium to final is 0.09m

    0.09----0------0.63

    What would be the total distance?
     
  15. Jun 5, 2010 #14
    I realize what I did now, the final equation should be

    Work by friction = M*(Normal)*(.63 - .63/7) = Efinal - Einitial

    because it's the change in distance. This works, thanks a ton!
     
  16. Jun 6, 2010 #15
    Don't forget the coefficient of kinetic friction:

    [tex]W_{fric}=\vec{F_{fric}}\cdot\vec{\Delta x}=-\mu_{kin}F_{normal}(\Delta x)=-\mu_{kin}mg(\Delta x)[/tex]
     
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