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MOTION EQUATION what am I missing here?

  1. Jan 5, 2007 #1
    a stone with masss of 0.62 kg is fired upwards and at an angle with SPEED OF 5.5ms (gravity lies at 9.8ms)

    I have kinetic energy at 0.5*0.62*5.5= 1.70500 j

    what is the gravitational potention energy PLEASE

    and how would I go ABOUT WORKING OUT THE HEIGHT HALF WAY THROUGH THE JOURNEY IF IT HAS A KINETIC ENERGY VALUE OF 2.3 JOULES AT THAT MOMENT EXACTLY???

    tHIS HAS THROWN ME !

    mANY THANKS IN ADVANCE AND PLEASE EXCUSE MY KEYBOARD , IT IS oN THE WAY OUT :(
     
  2. jcsd
  3. Jan 5, 2007 #2
    wrong section

    move topic please. this belongs in introductory physics help

    i dont understand that last bit about being halfway
     
  4. Jan 5, 2007 #3

    Integral

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    The gravitational potential depends on your choise of a coordinate system. Since you have not specified that, we cannot answer that question.

    Next, please show us your equation for kinetic energy. How does this compare to what you have written?
     
  5. Jan 5, 2007 #4
    sorry I wasnt aware thet it was in the wrong section...how do I go about moving it please :)
     
  6. Jan 5, 2007 #5

    cristo

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    A mentor needs to move it (and I see that Integral already has!) Have you thought about the points he mentioned above?
     
  7. Jan 5, 2007 #6
    Im really struggling with this, could you help ?
     
  8. Jan 5, 2007 #7
    a stone has a mass of 0.62 kg which is fired upwards at an angle to the ground and with a speed of 5.5 ms negative 1,air resistance is nil and the accelration ddue to gravity is 9.8 ms negative 2 .
     
  9. Jan 5, 2007 #8
    I need to work out the kinetic energy and the potential energy and in addition at another point which is half way through the flight the kinetic energy is set as 2.3 joules do I need to remove that from the potential energy to get my answer ??
     
  10. Jan 5, 2007 #9

    cristo

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    Well, as Integral says, you need to define your choice of coordinate system. (As in where the origin is). Is the stone being thrown from ground level or from a certain height?

    You say you want to calculate the potential energy: at what point do you want to know the potential energy?

    You need to clarify the specific details of the question!
     
  11. Jan 5, 2007 #10
    f= m a so 0.62kg * 5.5ms = 3.41 (momentum)
     
  12. Jan 5, 2007 #11
    sorry from the ground
     
  13. Jan 5, 2007 #12
    and returning to the ground with point b being mid flight (the 2.3j point)
     
  14. Jan 5, 2007 #13
    It doesn't sound like you need to know the momentum for this problem. You need to specify at what point you are calculating the potential energy. Where in the stone's flight are you calculating the potential energy?

    The kinetic energy equation is actually [itex]K = \frac{1}{2} mv^2[/itex], make sure you don't forget the "squared"!
     
    Last edited: Jan 5, 2007
  15. Jan 5, 2007 #14
    sorry will try again

    the kinetic as it laves the ground

    (Eg) gravitational potential energy max and min values

    and

    the kinetic energy of th stone at its maximum height (2.3 JOULES)
     
  16. Jan 5, 2007 #15

    HallsofIvy

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    No, you don't need to "work out the kinetic energy"- you are told that the kinetic energy is 2.3 Joules. While Integral is correct, that the potential energy depends upon the choice of coordinate system, I suspect that here the potential energy asked for is "relative to the initial position": it is mgh where m is the mass 0.62 kg, g is the acceleration due to gravity, 9.81 m/s2, and h is the height in meters above the initial position. The "physics law" you need here is conservation of energy: Since you are told the initial speed, you can calculate the initial kinetic energy. Of course, the potential energy, "relative to the initial position" is 0 so the kinetic energy is the total energy. You are told that the kinetic energy at the highest point is 2.3 Joules. Since total energy is a constant, the potential energy at the highest point must be the kinetic energy minus the kinetic energy at the highest point. From that use potential energy= mgh to find the height.
     
  17. Jan 5, 2007 #16
    So would the initial kinetic energy be 0.5*0.62*5.5 ???
     
  18. Jan 5, 2007 #17
    so would the height be 0.5*0.62*5.5 minus 2.3 ?
     
  19. Jan 5, 2007 #18
    No, kinetic energy is [itex]K = \frac{1}{2} m v^2[/itex], NOT [itex]\frac{1}{2} m v[/itex]. Square your velocity!
     
  20. Jan 5, 2007 #19
    or 0.62*5.5 = 3.41 minus 2.3 =1.1 metres ?
     
  21. Jan 5, 2007 #20
    ah 0.62*30.25= 18.75500
     
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