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Thread: Simple Problem(s)

  1. #1 Simple Problem(s) 
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    So I've got a rocket already in space with a mass of 100 kg and I want to know how much force will thrust it to 2 km meters per second. That's F = 100 x 200, right?

    Say I want to move something at the same speed, but its mass is only 10 g. Would that be F = 0.10 x 200?

    Does this formula only work if it's arranged with Newtons, kilograms and meters? Or could I multiply pounds and feet, then use a simple conversion to change the answer to Newtons?

    And one Newton is the force it requires to move a kilogram one meter in one second, right? Have I grasped this right?


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  3. #2 Re: Simple Problem(s) 
    Forum Ph.D. william's Avatar
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    Quote Originally Posted by GrantG
    So I've got a rocket already in space with a mass of 100 kg and I want to know how much force will thrust it to 2 km meters per second. That's F = 100 x 200, right?
    Not quite... F = ma, where 'a' is the acceleration. What you have is velocity (200 m/s).

    You might want to refer to the kinematic equations for constant acceleration:

    x<sub>f</sub>-x<sub>i</sub> = v<sub>i</sub>t + 0.5at<sup>2</sup>
    x<sub>f</sub>-x<sub>i</sub> = 0.5(v<sub>f</sub> + v<sub>i</sub>)t
    v<sub>f</sub> = v<sub>i</sub> + at
    (v<sub>f</sub>)<sup>2</sup> = (v<sub>i</sub>)<sup>2</sup> + 2a(x<sub>f</sub> - x<sub>i</sub>).

    Use one or more of these to find 'a.' You'll notice that, in addition to knowing the final velocity (if that's what you want to stick with), you'll also have to provide either the distance travelled or a time somehow.

    Say I want to move something at the same speed, but its mass is only 10 g. Would that be F = 0.10 x 200?
    Again, you need an acceleration - not a velocity. Otherwise you are almost correct - 10g = 0.01 kg.

    Does this formula only work if it's arranged with Newtons, kilograms and meters? Or could I multiply pounds and feet, then use a simple conversion to change the answer to Newtons?
    SI units are the easiest to work with (or at least that's what I think), but you can use other units as long as you are consistent throughout the calculation. E.g., lbs = slugs*feet/s<sup>2</sup>.

    And one Newton is the force it requires to move a kilogram one meter in one second, right? Have I grasped this right?
    Not quite...
    To rephrase your statement to make it accurate;
    One Newton is the force it requires to accelerate a kilogram to one meter per second-squared. You have to understand each term in the equation F = ma.

    If you play with the first kinematic equation above, you'll see that to move a kilogram a distance of one meter (assuming the initial velocity is zero), in one second, the acceleration turns out to be two meters per second-squared. This corresponds to a force of two Newtons.

    Using the same formula, and knowing the acceleration (2 m/s<sup>2</sup>), you can solve for the distance travelled in 2 seconds and find that it has travelled four meters. Etc., etc., etc.

    Cheers,
    william


    "... the polhode rolls without slipping on the herpolhode lying in the invariable plane."
    ~Footnote in Goldstein's Mechanics, 3rd ed. p. 202
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  4. #3  
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    no offence, and im only a high school student in a physics class, but wouldnt it be a hell of a lot easier to use the equation a=F/m
    i ripped this off of someone else's signature, but i felt that it equally applied to me.
    "Hammered Like A Blacksmith, Stoned Like a Mountain
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  5. #4  
    Forum Ph.D. william's Avatar
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    Quote Originally Posted by Perk
    no offence, and im only a high school student in a physics class, but wouldnt it be a hell of a lot easier to use the equation a=F/m
    Good question, but...

    Quote Originally Posted by GrantG
    ...I want to know how much force....
    Grant still didn't know the force (F). (That's ultimately what he was looking for - the force.) Using a=F/m, you have two unknowns and only one equation.

    You can use a=F/m if you know F and m.

    Cheers,
    william

    P.S. Also (for the hell of it...), F = dp/dt. This allows for a changing mass. Most of the time the mass is constant, but when it's not, you must use the derivative of momentum with respect to time.

    To see how this works;
    F = dp/dt = d(mv)/dt (and if m is constant -->) = m dv/dt = ma, (the form you're used to).
    "... the polhode rolls without slipping on the herpolhode lying in the invariable plane."
    ~Footnote in Goldstein's Mechanics, 3rd ed. p. 202
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