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A formula of a car using conservation of energy

  1. Nov 6, 2011 #1
    1. The problem statement, all variables and given/known data
    use conservation of energy to derive a formula for the speed of a car in terms of energy it has at the start. you have to use energy and height as the two variables


    2. Relevant equations



    3. The attempt at a solution
    i think it is kenetic energy x mgh
     
  2. jcsd
  3. Nov 6, 2011 #2

    PeterO

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    Does this problem perhaps relate to a roller coaster car?
     
  4. Nov 6, 2011 #3
    yes we did this roller coaster car problem in my physics class and i need to fill out an experiment sheet and this is one of the questions but i dont understand how to answer it
     
  5. Nov 6, 2011 #4

    PeterO

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    If you are allowed to ignore friction, then the sum of the potentail energy [mgh] and kinetic energy [0.5mv2] will be constant throughout.

    That means you will be able to find the kinetic energy at any time, by considering the height at that time. From that you can "undo" the kinetic energy formula to find the speed.
     
  6. Nov 6, 2011 #5
    im sorry but i still dont understand what would you do to find the formula
     
  7. Nov 6, 2011 #6

    PeterO

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    How was the initial energy of the car expressed?
     
  8. Nov 6, 2011 #7
    was it by gravity because it was at standstill then moved on a hill downwards
     
  9. Nov 6, 2011 #8

    PeterO

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    Actually I was asking a question rather than leading, but you have answered it any way.

    In the beginning, the car was at height H - how ever high the hill was. it will have had mgH of potential energy.

    when it is at any another height h, it will have descended an amount (H-h) and thus "lost mg(H-h) of potential energy.
    That will have been converted to kinetic energy, and from that kinetic energy you should be able to find the speed.

    As a test of your algebra .. if the Kinetic energy of the car was an amount, A, what will the speed of the car be. [I used A as I needed a pronumeral for the amount, and didn't want to use E unless you confused it with a specific energy]
     
  10. Nov 6, 2011 #9
    so i will use kinetic energy formula= height and then solve for speed

    so it would be .5x.05796xV^2=.09
     
  11. Nov 6, 2011 #10

    PeterO

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    No, kinetic energy = loss of potential energy, which involves height, and mass and g [9.8].

    Where did you get those values .05796 and .09 from?
     
  12. Nov 6, 2011 #11
    sorry those were from the data .09 was height and .057 was mass

    so how would you put the formula together. with what symbols would you use. would you use GxMxH-h from what you said. i dont know how to put it together. if you could explain what i would use then it would mean alot even though you already helped me alot
     
  13. Nov 6, 2011 #12

    PeterO

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    Repeating an earlier request.

    If the kinetic energy has value "A" what is the speed of the mass?
     
  14. Nov 6, 2011 #13

    PeterO

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    Also that height of .09.

    What measurement was .09? The height of the start point? The height of the mass at a time? The height "below" the starting point?
     
  15. Nov 6, 2011 #14
    A=.5xMxV^2

    i dont know if this is right. is this how i would put the formula together.

    or would it be A=MGH but how would you find velocity

    im sorry but im confused
    the height at the start point
     
  16. Nov 6, 2011 #15
    peter please if you could just give me the formula. i need to answer this question by tommorow and if i dont i will fail. i will ask my teacher to explain it but i cant ask him now. please will you give me the formula
     
  17. Nov 6, 2011 #16

    PeterO

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    I am in Australia - not sure what time it is where you are.

    A=.5xMxV^2

    is correct, and this leads to V = √(2A/M).

    You can use "change in Potential energy" to find how much energy you have.

    IF that was mgΔh, your speed would be given by V = √ (2mgΔh/M)

    However, you are not using m for mass, and Δh will be worked out from the h you are at and the height you began at.
     
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