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Newton's Laws, Force and Kinematics

  1. Jul 27, 2009 #1
    1. The problem statement, all variables and given/known data
    A particle is traveling in a straight line at a constant speed of 24.5 m/s. Suddenly, a constant force of 14.3 N acts on it, bringing it to a stop in a distance of 52.9 m.

    a. Determine the time it takes for the particle to come to a stop.

    b. What is its mass?


    2. Relevant equations

    1. vx2 = v0x2 +2a[tex]\Delta[/tex]x

    2. [tex]\Delta[/tex]x = v0t + 0.5at2

    3. F=ma

    3. The attempt at a solution

    I know how to solve part a. I use equation 1 to find a which is 5.673 m/s^2
    I use that value of a in equation 2 to find t which is 4.318 s
    This is all correct.

    I am having trouble with part b - to find mass.
    The only equation I have to find mass is f=ma
    F is 14.3N given, a as calculated is 4.318 which gives m as 3.311 kgs which is incorrect.
    The correct answer is 2.52 kg

    So I thought that for F=ma, F must be the resultant force of some sort.
    So I subtracted 24.5-14.3 (which I am sure is not a correct step to do) gives me 10.2

    And that divided by a (4.318) gives me m as 2.362 which is close but not correct.

    Any other thoughts? Thanks so much


    EDIT: Aarggghh.. stupid me
    I am diving by t (4.318) instead of the a I found (5.673).
    Can't believe I spent so much time on this stupid stupid mistake.
     
  2. jcsd
  3. Jul 27, 2009 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi skydiver! Welcome to PF! :smile:

    I'm glad you've sorted it out! :biggrin:

    Just for the record, there's another way of finding the mass …

    just calculate the work done and then use the work-energy theorem … work done = change in energy. :wink:
     
  4. Jul 27, 2009 #3
    Re: Welcome to PF!

    The really beautiful thing is, that the first equation he posted:
    [tex]v_f^2=v_0^2+2ad[/tex] is completely equivalent.
    Multiplying by a factor of [tex]\tfrac{1}{2} m[/tex] provides:

    [tex]\tfrac{1}{2} mv_f^2=\tfrac{1}{2} mv_0^2+Fd[/tex]

    (The dot operator is included in the way the equation is defined, since the [tex]a[/tex] term refers to the acceleration in the direction of the distance traveled.)
     
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