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Homework Help: 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
     
  22. Jan 5, 2007 #21

    cristo

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    No, the formula for kinetic energy is KE=mv2/2.

    You appear to just be throwing numbers together now. Think of the physical law which needs to hold, namely conservation of energy. You can calculate the kinetic energy at the beginning, which is the total energy of the system (since there is no PE at the initial point). The conservation of energy states that the sum of kinetic and potential energy of the system, at any one point, must equal the total energy. Since you know the kinetic energy at the point required, then you can calculate the potential energy, and thus the height.

    Do you know an equation relating potential energy and height?
     
  23. Jan 5, 2007 #22
    then minus the 3.2 ? for the height ??
     
  24. Jan 5, 2007 #23
    0.62*30.25= 18.75500 please tell me tht is correct?
     
  25. Jan 5, 2007 #24
    You probably can't solve physics problems by batting around in the dark. You know that all the forces acting in this problem are conservative (no energy is lost to friction or air resistance or so on) so you can use conservation of energy to figure it out, which means total energy (E) is conserved throughout the ball's flight from point 1 to point 2:

    [tex]
    E_1 = E_2
    [/tex]

    since E = K+U

    [tex]
    K_1 + U_1 = K_2 + U_2
    [/tex]

    Anywhere in the ball's path you will find that the TOTAL mechanical energy is always the same (even though the values U and K change, they always add up to the same number). So, you can pick two points to use: for this problem, it would be best to pick when the ball is just leaving the ground, and when it is halfway through its flight (since you are given information about these two points).

    You are given the values to calculate [itex]K_1, U_1[/itex] in the problem, and [itex]K_2[/itex] is given explicitly. That leaves only one unknown variable, [itex]U_2[/itex], which, once solved for, you can use to find the height!
     
    Last edited: Jan 5, 2007
  26. Jan 5, 2007 #25

    cristo

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    Please read my post, and answer my questions. You will need to post equation, (with algebraic symbols in) instead of strings of numbers without explanation of where they come from, or else noone can help you!
     
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