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Kinetic Energy and height unknown

  1. Mar 8, 2007 #1
    1. The problem statement, all variables and given/known data
    Determine the KE of a 2000 kg car with speed = 22.3 m/s.

    2. Relevant equations
    E = 0.5 x 2000 x 22.3^2= 497290 Joules or 497 KJ.

    3. The attempt at a solution

    The car is lifted vertically and dropped from rest. Find the height from which it was released if v before hitting the ground = 22.3m/s.

    Here I don't know what to do...

  2. jcsd
  3. Mar 8, 2007 #2
    Hint: Potential energy
  4. Mar 8, 2007 #3
    Another hint:Think of total energy, what can you say about the total energy of the car?
  5. Mar 8, 2007 #4
    Is this for the first or the second question?
  6. Mar 8, 2007 #5
    both hints , are for the second part, where you need to know the height.
  7. Mar 8, 2007 #6
    PE = m x g x h
    m = 2000Kg
    g = 9.8 m/s^2
    h = the unknown
    PE = ? is it related to v?
  8. Mar 8, 2007 #7
    the potential enegry at the height h, should be equale to the potential energy of the car as it reach the ground..
    Et is conserved..
    can you continue now ?
  9. Mar 8, 2007 #8
    Is the total energy the value I found in the first part? 497 KJ?
    So I divide 497 by 2 to find Ep?
    And then I can find h...

    Is this correct?
  10. Mar 9, 2007 #9
    I think ziad meant to say, "the potential enegry at the height h, should be equal to the kinetic energy of the car as it reach the ground."

    In general, for a particle moving in a gravitational field not affected by forces of friction and such, the total energy will remain a constant, and it will be equal to the sum of its (gravitational) potential and kinetic energies.
    Last edited: Mar 9, 2007
  11. Mar 9, 2007 #10
    I am completely confused... Could you show me what you would do so that I can understand the process?
    Do we still use: PE = m x g x h and Et=Ek+Ep ?

    Thank you...
  12. Mar 9, 2007 #11

    Et, at all times, is a constant.

    When the car is at a height h (which you need to find out), its potential energy(Ep) is mgh, but it is at rest, so Ek = ?

    But when the car hits the ground, it does so with some velocity, and you already know what it's kinetic energy is at the velocity. If you take the ground to be the 0 of the height, Ep at the bottom = ?

    Et(at h) = Et(at 0), solve for h.
  13. Mar 9, 2007 #12
    Here's what I did, if it's not right, then I have no clue...

    h = 1/2mv^2/mg=1/2v^2/g
    = (22.3^2 x 0.5)/9.8 = 25.4 meters.
  14. Mar 9, 2007 #13
    That's right.
  15. Mar 9, 2007 #14
    Thank you!
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