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Net Force of object changing velocity

  1. Mar 27, 2008 #1
    [SOLVED] Net Force of object changing velocity

    Hi, I'm studying for a test, so there might be a few of these in the next hour or so.

    1. The problem statement, all variables and given/known data
    A supertanker of mass 1.0 X 10[tex]^{8}[/tex] kg travels 3.5km, reaching a speed of 4.1km/h from rest. What was the magnitude of the unbalanced force acting on it?

    2. Relevant equations
    Fnet = ma
    v[tex]^{2}_{2}[/tex] = v[tex]^{2}_{1}[/tex] + 2a[tex]\Delta[/tex]d


    3. The attempt at a solution
    I tried (4.1km/h)[tex]^2[/tex] = 2a(3.5km)
    [tex]\frac{16.81km/h^2}{3.5km}[/tex] = 2a
    4.80km = 2a
    2.4km/h = a

    and then I tried converting it by dividing it by 60 twice, and sliding to the right 3 decimal places, then inserting it into the Fnet = ma.

    I'm really stuck, and that's not good if I have a unit test tomorrow :frown:
     
    Last edited: Mar 27, 2008
  2. jcsd
  3. Mar 27, 2008 #2

    rock.freak667

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    When using F=ma, you need m to be in kg and a to be in m/s^2
    So, convert the velocities into m/s and the distance into m
     
  4. Mar 27, 2008 #3
    I'll try it again but instead beforehand.. I did it after, though, does that make a difference?
    EDIT: still not working out though.
     
    Last edited: Mar 27, 2008
  5. Mar 27, 2008 #4

    rock.freak667

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    Not really but it seems correct that [itex]a=2.4kmh^{-2}[/itex]. Just multiply by 1000 and divide by 3600 and it should be correct.
     
    Last edited: Mar 27, 2008
  6. Mar 27, 2008 #5
    The answer in the book is 1.9 * 10[tex]^4[/tex]

    By multiplying 2.4 times 1000 and dividing by 3600 gives me 0.67. Still not correct. :S
     
  7. Mar 27, 2008 #6

    rock.freak667

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    Divide by 3600^2 . Sorry,told you to do the wrong thing.
     
  8. Mar 27, 2008 #7
    OK Thanks.
     
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