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Work-Energy Theorum: Spring potential energy vs Kinetic Energy

  1. Oct 10, 2009 #1
    [SOLVED] Work-Energy Theorum: Spring potential energy vs Kinetic Energy

    1. The problem statement, all variables and given/known data

    A 1350-kg car rolling on a horizontal surface has a speed v = 40 km/h when it strikes a horizontal coiled spring and is brought to rest in a distance of 2.5 m. What is the spring constant of the spring? Ignore Friction and assume spring is mass-less.

    2. Relevant equations
    [tex] W = \Delta E[/tex]
    [tex] E_{pspring} = \frac{1}{2}(kx^2) [/tex]
    [tex] E_k = \frac{1}{2}(mv^2) [/tex]


    3. The attempt at a solution

    First right off the bat, i converted 40 km/h to its m/s equivalent of aprox. 11.11 m/s

    i state the law of conservation of energy: Energy before = Energy after

    Therefore:

    [tex]
    E_k = E_{pspring}
    \frac{1}{2}(mv^2) = \frac{1}{2}(kx^2)
    [/tex]

    then i isolate k

    [tex] k = \frac{-mv^2}{x^2} [/tex]

    now heres the issue, is x negative? because the displacement is against the direction of motion?
    and 2.5m = x, (-2.5)^2 gives me a answer of 4266 Nm
    but -(2.5)^2 is entirely different.. This has been a long lasting math issue for me.

    And what if x is positive?

    i know k MUST be positive right?
     
    Last edited: Oct 11, 2009
  2. jcsd
  3. Oct 10, 2009 #2

    rl.bhat

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    Homework Helper

    (2.5)^2 is correct. There is no negative energy in the nature.
     
  4. Oct 10, 2009 #3

    Delphi51

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    Homework Helper

    There is no minus sign in mv^2 = kx^2
    or in k = mv^2/x^2. No way you can get k negative!
    The minus sign in F = -kx is supposed to help keep track of the fact that the force of the spring is opposite to the direction of stretch but it does seem to have a habit of getting in the way. k is ALWAYS positive.
     
  5. Oct 11, 2009 #4
    Thanks, the negative sign on mv^2 was an algebra error... Thanks for the clarification guys!
     
  6. May 18, 2010 #5
    Your attempt is correct but you missed somethingthat for a spring if you take natural length as the datum, the force on change in length is given as:
    [tex] \vec{F}= -k \vec{x} [/tex]

    and hence work done by a spring against external forces
    [tex] W_{s}=\int\vec{F}\vec{.dx} [/tex]
    over the required limits

    in our case the answer is
    [tex]W_{s}=-\frac{kx^{2}}{2}[/tex]
    as
    [tex]W_{s}=\Delta E[/tex]
    [tex]\Delta E=-\frac{mv^{2}}{2}[/tex]
    the change part was where you lost it all...the KE FELL TO ZERO. HENCE A NEGATIVE CHANGE.
     
    Last edited: May 18, 2010
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