1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    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.)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook