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Which formulae?

  1. Jul 3, 2007 #1
    which formulae???

    hi there

    i have an exam soon and i have the question that follows.

    A body of mass of 2kg is travelling with a speed of 5m/s. If a force of 5N acts upon it due to friction, how much distance will it travel?

    does anyone know which formula i need to use?? i am really stumped.

    many thanks in advance
  2. jcsd
  3. Jul 3, 2007 #2


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    Welcome to PF, nobby. Note that for homework and coursework questions we require that you show some work before we can help you.

    That said, I'll give you one hint: try using the relationship between work and kinetic energy.
  4. Jul 3, 2007 #3
    Cristo you little beauty

    once i got both formula on paper it all made sense. thank you very much

  5. Jul 3, 2007 #4


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    You're welcome!
  6. Jul 3, 2007 #5
    cristo maybe you could help with another question.

    the question is....An aircraft weighing 6400 pounds lands at a speed of 10ft/sec and stops in 10 seconds. What force was generated by the brakes? (assuming gravity as 32ft/sec^2)

    when i worked this out I converted the pounds and feet/sec to kg and m/s, was this the right thing to do or should i just have carried out the workings out in the original terms??? I again used the kinetic energy and work done formula and came up with 442.9N.

    my working out is thus.

    mass = 6400lb x 0.453 (to go to kg) = 2903kg
    velocity = 10ft/sec x 0.3048 (to go to m/s) = 3.048m/s
    time = 10 sec
    displacement = 100ft (i assumed this as 10ft/sec for 10sec) x 0.3048 (for metres) = 30.48m

    Ke = 1/2mV^2
    Ke = 1/2 x 2903 x 3.048^2
    Ke = 1451.5 x 9.3
    Ke = 13484.9

    work done = force x distance
    transposed is force = work/distance
    force = 13484.9/30.48
    force = 442.9N

    does this sound right to anyone??? if anyone has any feed back i would love to hear from you

    many thanks in advance
  7. Jul 3, 2007 #6


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    It doesn't really matter in which units you work. I prefer SI, but as long as you are consequent about it, it doesn't really matter (e.g. divide meters by meters and not by feet).

    Your answer looks fine. Note that you have implicitly used here, that all the kinetic energy is converted into work (the kinetic energy at the end is zero), as the plane slows to a stop. This would not have been the case if for example, it just braked to 1 m/s and then rolled out by friction -- so it's an important assumption.
  8. Jul 3, 2007 #7
    thank you for your feedback and comments they are much appreciated.
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