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Conservation of energy

  1. Jun 15, 2008 #1
    a 1500kg car slams on the brakes on an icy road with a coefficient of kinetic friction of .05. the car has initial velocity before applying the brakes.find the distance requires for the car to stop completely. use the principles of conservation of energy.
  2. jcsd
  3. Jun 15, 2008 #2
    You need to show your work.
    Start with the definitions. The mechanical energy lost is [tex]f \Delta s=\Delta E_{therm}[/tex], where f is the magnitude of the friction force(what is the definition of kinetic friction?), and [tex]\Delta s[/tex] is the distance that you are trying to find.

    By the work-energy theorem, [tex]W_{ext}=\Delta E_{mech}+\Delta E_{therm}[/tex]. Now, solve for [tex]\Delta s[/tex]
  4. Jun 15, 2008 #3
    yea i need to show work... but what is delta mech and delta therm i dont know how to set up the problem..
  5. Jun 15, 2008 #4
    I just gave you the definition of [tex]\Delta E_{therm{[/tex]; [tex]\Delta E_{mech}[/tex] is just the mechanical energy i.e. the initial kinetic energy. Open up your textbook and look at the definitions.
  6. Jun 15, 2008 #5
    im lost really i just dont get it..would you kbe kind enough and plug in nummbers
  7. Jun 15, 2008 #6
    Absolutely not. You haven't done anything. As for the answer, it's very easy to see you just have to do it yourself. What is the definition of kinetic energy? What is f?
  8. Jun 15, 2008 #7
    kinetic energy is (1/2) M V^2 and F= friction
  9. Jun 15, 2008 #8
    No, I meant force of friction. What is the definition of force of friction? After you've done that, then you can plug it in the equation that I told you:

    [tex]W_{ext}=\Delta E_{mech}+\Delta E_{therm}[/tex]
    You know,
    (1) [tex] f \Delta s=\Delta E_{therm} [/tex]
    (2) [tex]W_{ext}=\Delta U + \Delta K + f\Delta s[/tex]
    (2) As you've just said, [tex] \Delta K =\frac{1}{2}mv^2[/tex]

    Now, ask yourself, what must [tex]W_{ext}[/tex] be? Then write the definition of f. Then find s.
    Last edited: Jun 15, 2008
  10. Jun 15, 2008 #9
    this is wat i have

    (1/2)(1500)v^2+.05 delta s =.05 delta s
  11. Jun 15, 2008 #10
    No.You are given the coefficient of friction,, which is denoted as "mu" [tex]\mu[/tex]. What is the friction force? [tex]f=\mu*blank[/tex] What is blank? It's a definition; look it up in your book.

    Where did you get the terms on the right-hand side? What is [tex]W_{ext}[/tex]? Hint: There are no conservative forces and no external forces acting on the system, then W=?
  12. Jun 15, 2008 #11
    its equal to N which is mass times gravity ....i been on this problem for almost 3 3hours long and i got like 5 more to go and its due tomorrow plz plz plz plz
  13. Jun 15, 2008 #12
    Ok, now plug it back into the equation, we get

    [tex]W=\frac{1}{2}mv^2+\mu mg \Delta s[/tex]

    You're almost there. What must W be? Then you can solve for [tex]\Delta s[/tex]
  14. Jun 15, 2008 #13
    w must be the same as (1/2)mv^2 + "mu"mg
  15. Jun 15, 2008 #14
    No. If there are no conservative forces doing work and no external forces acting on the system, then W=?
  16. Jun 15, 2008 #15
    it should equal 0
  17. Jun 15, 2008 #16
    it should equal 0
  18. Jun 15, 2008 #17
    Yes! Now, solve for the distance, [tex] \Delta s [/tex]
  19. Jun 15, 2008 #18
    i got 1.02m i dont think it my answer makes sense
  20. Jun 15, 2008 #19
    What do you mean it doesn't make sense? Show your work.
  21. Jun 15, 2008 #20
    alright this is wat i put in.... (1/2)(1500)(cos180)^2 divided by (.05)(1500)(9.8)
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