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Potential/conservation of energy

  1. Jul 17, 2006 #1
    i got part of of this problem but b, c, and d I can't get. For c i think i am supposed to use k=1/2mv^2 but i dont know for sure? please help.

    During a rockslide, a 340 kg rock slides from rest down a hillside that is 500 m long and 300 m high. The coefficient of kinetic friction between the rock and the hill surface is 0.28.
    (a) If the gravitational potential energy U of the rock-Earth system is set to zero at the bottom of the hill, what is the value of U just before the slide?
    (b) How much energy is transferred to thermal energy during the slide?
    ? J
    (c) What is the kinetic energy of the rock as it reaches the bottom of the hill?
    ? J
    (d) What is its speed then?
    ? m/s
  2. jcsd
  3. Jul 17, 2006 #2

    is the ans for (b) 466301 J ?
  4. Jul 17, 2006 #3

    Doc Al

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    Staff: Mentor

    Two (equivalent) ways to approach this:
    (1) via energy: How much work is done against friction?
    (2) via dynamics: What's the acceleration of the rock?
  5. Jul 17, 2006 #4
    remember that no energy can be lost !

    The amount of energy at the beginning of the movement equals that after the movement. Thus from the moment at rest till tdown at the hill:

    [tex] E_{pot} = E_{kinetic} + E_{friction}[/tex]
    [tex] mgh = \frac {1} {2} \ mv^{2} + E_{friction}[/tex]
    Last edited: Jul 17, 2006
  6. Jul 17, 2006 #5
    thecanswer for b was not correct
  7. Jul 17, 2006 #6
    i dont have a velicity to work with
  8. Jul 17, 2006 #7
    that's what you need to calculate from this energy balance ;) For B [tex]E_{friction}[/tex] is asked.

    [tex] E_{friction} = F_{friction} * s [/tex]
    Last edited: Jul 17, 2006
  9. Jul 17, 2006 #8


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    Staff Emeritus
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    Gold Member

    For b: What exactly is creating the thermal energy? I'll give you a hint... it starts with 'frictio' and ends with 'n' :wink:

    Knowing this, is there any good way to calculate the total thermal energy created by the friction?

    c: Once you've gotten b, what types of energy is the potential energy being converted to? Then just keep in mind conservation of energy, and you should be able to get the answer

    d: Given the kinetic energy, velocity is easy to calculate
  10. Jul 17, 2006 #9
    not even close? :cry:
  11. Jul 17, 2006 #10
    i dont know the answer to b yet
  12. Jul 17, 2006 #11
    would i just use the coefficient of kinetic friction times the normal force for the enery of friction
  13. Jul 17, 2006 #12
    yes, but you need the work done by this force as you need to know the energy of friction that is lost as heat.

    Hint: [tex]F_{N}[/tex] is not [tex]F_{g}[/tex]
  14. Jul 17, 2006 #13


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    Yes. potential energy = mgh= (340)(300)(9.8) approximately.

    A complicated calculation. You know the coefficient of friction and you can calculate the component of weight perpendicular to the hill (better draw some triangles to analyze that), then multiply by the coefficient of friction, 0.28, to get the actual friction force. Multiply the friction force by the length of the slope (use the Pythagorean theorem to get that) to find the energy expended in sliding down the slope. Of course, that work becomes thermal energy (heat) in the hillside.

    Added later: No! It's not at all complicated. The energy used up in friction (which becomes thermal energy) is just the coefficient of friction times the Potential energy! That's remarkable. I had to do the calculation for any slope before I believed it.

    You know the potential energy at the top of the hill and, since the rock was sitting still, it had no kinetic energy so that (your answer to (a)) is the total energy. Since the rock has potential energy 0 at the bottom of the hill, part of the total energy has gone into thermal energy (part (c)) and what is left is kinetic energy.

    Kinetic energy is (1/2)m v2. You are given m and you know the kinetic energy from (c). Solve for v.
    Last edited: Jul 17, 2006
  15. Jul 17, 2006 #14
    oookk. I got it now. thanks to all who helped
  16. Jul 17, 2006 #15
    i didn't get it
  17. Jul 17, 2006 #16
    lol, what you didn't get :smile:
  18. Jul 17, 2006 #17
    for b i did 340 kg* 9.8 (for the force perpendicular to the hill) times the coefficient of kinetic friction, .28 then I muiltiplyed that by the lengght of the hill.583.1

    i got 544008.976 could this be right?
    Last edited: Jul 17, 2006
  19. Jul 17, 2006 #18
    we have a hill here with a certain angle. The vector of the normal force stand in a right angle at the surface.

    --> you need to make 2 vectors of the gravitational force: 1 in the direetion of the movement and 1 at a right angle at the surface.

    The angle can be calculate by the length of the hill and the highth use goniometry.

    Thus: [tex] F_{n} = F_{g} * cos \alpha [/tex]
    Last edited: Jul 17, 2006
  20. Jul 17, 2006 #19
    so one vector directed down the hill at 9.8 and one vector directed perpendicular to the hil at 3332
  21. Jul 17, 2006 #20
    i got an angle of 59 degress tan-1(500/300). then i did 9.8cos59=5.04
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