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Kinetic energy

  1. Jun 12, 2008 #1
    1. The problem statement, all variables and given/known data
    A 3.00kg object has velocity 7.00j m/s. Then, a total force 12.0i N acts on the object for 5.00s.
    Find the final kinetic energy from 1/2mvf^2 = 1/2mvf . vf


    2. Relevant equations



    3. The attempt at a solution
    i got 1/2 * 3 * (20i+7j)(20i+7j)
    then 600j^2 + 210ij + 73.5j^2

    but the answer is 673.5 J, why?
    since j.j =i.i=1 and i.j=0?
     
    Last edited: Jun 12, 2008
  2. jcsd
  3. Jun 12, 2008 #2

    Kurdt

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  4. Jun 12, 2008 #3
    !? do i need to write the whole of that!!??

    600j^2 + 210ij + 73.5j^2
    600j.j + 210 i . j + 73.5j.j
    600(1) + 210 (0) + 73.5(1)
    is this alright
     
  5. Jun 13, 2008 #4

    Kurdt

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    No thats not how a dot product works. You don't expand the brackets.

    [tex] \mathbf{a} \cdot \mathbf{b} = a_xb_x + a_yb_y [/tex]

    For vectors:

    [tex] \mathbf{a} = (a_x,a_y) [/tex]
    [tex] \mathbf{b} = (b_x,b_y) [/tex]
     
  6. Jun 13, 2008 #5

    got it, a.b = axbx+ayby
    a.b= 20i*20i+7j*7j= 449
     
    Last edited: Jun 13, 2008
  7. Jun 13, 2008 #6

    Kurdt

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    A dot product gives a scalar answer and thats why it is sometimes called a scalar product. Drop the i's and j's and add the numbers together.
     
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