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A Couple of Basic Questions about Mechanics

  1. Nov 16, 2011 #1
    I'm writing a paper on a propulsion device and I could use some F/P ratio comparisons, like typical values for Newtons per Watt for propeller driven, jet driven, and rocket driven craft.

    Also I ran in to a confusing point involving work and kinetic energy. if kinetic energy is 1/2 m v^2, then say a 1kg mass that is traveling at 1m/s, then according to the formula the mass has a kinetic energy of 0.5 joules.

    Now let say the mass was accelerated from a velocity of zero to the velocity of 1m/s, by a force of 1 newton applied for one second. The definition of work state that the work applied to a mass increases the kinetic energy of that mass. So the 1 newton force applied for one second, resulted mass moving one 1 meter. The formula for work states that force times distance equals the energy added to the system, Thus 1 newton times 1 meter should equal one joule of energy added to the mass.

    This is how I get two different answers for the kinetic energy of a 1kg mass that is moving at 1m/s.

    Could someone point out my mistake in this please.

    Best regards,

    FredB
     
  2. jcsd
  3. Nov 16, 2011 #2
    Your mistake is that a 1 N force applied for one second means that the body travels one meter. Your "jerk" (rate of change of acceleration) for 0 < t < 1, is 1. So integrating that 3 times you get:
    [tex] s(t) = \frac{x^3}{3} [/tex] for 0 < t < 1

    So the displacement is s(1) = [itex] \frac{1}{3} [/itex]
    Now an easier way to get work from this:
    [tex] Work = \delta E_k [/tex]

    where your speed function is:
    [tex] v(t) = \frac{x^2}{2} [/tex]

    this should clear things up.
     
    Last edited: Nov 16, 2011
  4. Nov 16, 2011 #3
    I'm still confused.

    Maybe could you point me to a good article on the subject. Wikipedia is what has confused me.

    Then again maybe I can just ignore the N*m approach and use the delta Ek method instead.

    I could still use some ball park figures for F/P ratios for propeller planes. jets, and rockets...
     
    Last edited: Nov 16, 2011
  5. Nov 16, 2011 #4
    All you need to know is that [itex]a(t) = v'(t) = s''(t)[/itex] and that [itex] s(t) = \int v(t) dt = \int \int a(t) dt^2[/itex].

    And for work, there are a couple of formulas you can use.

    [tex] W = \int F \cdot dx = \int F \cdot v dt [/tex]
    [tex] W = \Delta E_{k} [/tex]

    In terms of a good article, I know of none but i'll look. Best description i've ever seen of work is in the Feynman Lectures - Volume 1. (Maybe because I'm a huge sucker for Feynman :P)

    EDIT: Maybe try this page: http://physics.info/work/
     
    Last edited: Nov 16, 2011
  6. Nov 16, 2011 #5
    The rate of change of velocity is called "acceleration". Jerk is the rate of change of acceleration.
    In the example described in the OP there is a constant acceleration 1m/s^2.
    The distance traveled in 1 s is 0.5 m and the work is 0.5 J, equal to the change in kinetic energy.

    To the OP: just look up uniform accelerated motion.
     
  7. Nov 16, 2011 #6

    AlephZero

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    IF you are still confused, igmore posts #2 and #4, which are also confused and/or wrong.

    To repeat what #5 said, in slightly different words:

    The velocity increases linearly from 0 to 1 m/s over 1 second.
    So the average velocity over the second is (0 + 1)/2 = 0.5 m/s
    Distance traveled = average speed x time = 0.5 meters.
    The work done = force x distance = 0.5 joules.
    And that is the same as the kinetic energy = [itex]mv^2/2[/itex].
     
  8. Nov 16, 2011 #7
    I see.

    So I erred in presuming that 1N applied to 1kg for one second would result in a displacement of 1m. It actually results in a displacement of 0.5m, and 1N acting through a distance of 1m would increase the kinetic energy by 1 joule it would just take longer than 1 second if the origninal velocity was zero.

    Thanks, that's a relief. At least I know where to look now.

    the calculus text I have on hand is rather poor. Barely a paragraph about the dot product of N*m. It's single variable calculus for non-science/engineering majors.

    Feyman, I'm a fan. I'll see if I can find any of his basic physic lectures that were taped.
     
  9. Nov 16, 2011 #8
    #4 is correct. Some of #2 is incorrect.
     
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