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Newton's Laws magnitude of the force

  1. Oct 14, 2009 #1
    First post =]
    Requesting guidance with this particular problem:

    1. The problem statement, all variables and given/known data
    A 1.20 kg object is moving in the x direction at 14.4 m/s. Just 3.31 s later, it is moving at 28.8 m/s at 34.0 degrees to the x axis.

    a.) What is the magnitude of the force applied during this time?

    b.) What is the direction of the force applied during this time?

    2. Relevant equations

    3. The attempt at a solution
    I first organized all my givens, that is:
    mObject = 1.20kg
    vObject = 14.4m/s
    t = 3.31s
    v2Object = 28.8m/s 34 degrees to x axis

    I established a coordinate system so that positive would be ^ and -> , and negative would be the opposites. I then drew the corresponding vectors (14.4m/s and 28.8m/s at the angle).

    Then I broke down the angled vector into its x and y components:

    v2x = 28.8cos34 = 23.88m/s
    v2y = 28.8sin34 = 16.1m/s

    Here's where I get uncomfortable with my work:

    I know Fnet = ma

    So F = 1.2(v/t)
    = 1.2(v2-v1 / t)
    = 1.2kg((23.88m/s - 14.4m/s) / 3.31s)
    = 3.44

    I'm aware I completely neglected the y component, and I've been having trouble finding out where to fit it really. I'd love a push in the right direction if someone wouldn't mind - thanks!
  2. jcsd
  3. Oct 15, 2009 #2


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    Homework Helper

    Hi mysteek, welcome to PF.
    Force is defined rate of change of momentum.
    Initial momentum Px is along the x direction. Resolve the final momentum into Px' and Py'.
    Find the magnitude of change in momentum and hence the force.
  4. Oct 15, 2009 #3
    Ah, momentum, brings back memories.

    So then:

    Px(initial) = mass of object * initial velocity
    = 1.2kg(14.4m/s)
    = 17.28

    Px(final) = mass of object * x component final velocity
    = 1.2kg(23.88m/s)
    = 28.656

    Py(final) = mass of object * y component final velocity
    = 1.2kg(16.1m/s)
    = 19.32

    Now that I have the components, I can compute the change in momentum, thus providing me with force:

    Fnet = Change in P

    Change in Px = P(final) - P(initial)
    = 28.656 - 17.28
    = 11.376 N

    Change in Py = 19.32 N

    Magnitude of change in momentum would be:

    Sqrt(11.376^2 + 19.32^2)
    = 22.4 N

    Am I missing something? I didn't use the 3.31 seconds of time. Was ''time'' in the question just for the sake of throwing me off?
  5. Oct 15, 2009 #4


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    To get F divide change in momentum by time.
  6. Oct 15, 2009 #5
    22.4 N would be the magnitude of the change of momentum, if I were to divide this by time I would get

    22.4 / 3.31s

    = 6.77 N

    Oh, right! That makes perfect sense all the sudden, a bit more reading had me discover that momentum is in Newton seconds, so if I divided by time, seconds would cancel and I'd be left with newtons, aka force.

    At least I think my logic is correct. I'd like to think I understand the concept now.
  7. Oct 15, 2009 #6
    Sorry for double posting - for some reason my browser isn't letting me click edit on my last post :(

    When finding the direction of the force applied, could I use sine law?


    6.77N is the magnitude of the change of momentum, thus it would represent the hypotenuse of a right angle triangle, and because this vector is the hypotenuse, I can use:

    (19.32 is the change in momentum in the y direction, aka opposite to the angle we need)

    Sinx = opposite/hypotenuse
    = 19.32 / 6.77

    I get a math error when trying to find x. I assume I'm not using the correct numbers?

    Crap, attempt number 2:

    I think that the magnitude of the change in momentum has nothing to do with this particular section of the problem.

    I found the hypotenuse earlier, did I not?

    Sqrt(11.376^2 + 19.32^2)
    = 22.4 N s

    This value is needed to keep units consistent, since the x and y components are also in N s.

    Thus, I think I would use my same approach with this value?

    sinx = o/h
    = 19.32 / 22.4
    x = 59.6 degrees?
    Last edited: Oct 15, 2009
  8. Oct 15, 2009 #7


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    tanθ = Py/Px
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