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Newton's second law and direction of force

  1. Oct 11, 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 22.1 m/s. Suddenly, a constant force of 12.6 N acts on it, bringing it to a stop in a distance of 55.3 m.


    2. Relevant equations



    3. The attempt at a solution

    (a) What is the direction of the force?
    perpendicular to the direction of the particle's motion
    *opposite the direction of the particle's motion
    the same as the direction of the particle's motion
    none of the above

    (b) Determine the time it takes for the particle to come to a stop.
    ___ s

    (c) What is its mass?
    ___ kg

    F=ma
    [tex]V_f=0[/tex]
    [tex]V_f ^2 = V_i ^2 + 2a(X-X_i)[/tex]
    [tex]0^2=(22.1m/s)^2 + 2a(55.3m)[/tex]
    [tex]0=(22.1m/s)+at[/tex]
     
  2. jcsd
  3. Oct 11, 2009 #2

    rock.freak667

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    Homework Helper


    Yes so the deceleration is ? And the time taken is?
     
  4. Oct 11, 2009 #3
    [tex]0=22.1m/s^2+2a(55.3)[/tex]
    [tex]-110.6a=488.41m/s[/tex]
    [tex]a=-4.416m/s[/tex]

    [tex]0=22.1m/s^2+(-4.416m/s)t[/tex]
    [tex]22.1m/s^2/4.416m/s=5.004s[/tex]
    [tex]t=5.004s[/tex]

    Is this right?

    How do I get mass from this? F=ma right, so m=F/a or 12.6=m(-4.416)
    or m=12.6/-4.416. But it doesn't seem to work...
     
    Last edited: Oct 11, 2009
  5. Oct 12, 2009 #4

    rl.bhat

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    Homework Helper

    c) Use 1/2*m*v^2 = F*d
     
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