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Velocity from force/time graph

  1. Mar 26, 2015 #1
    Hey all, i've been attempting this all week and can't get my head around what the proper way to calculate the answers for this query is so...

    1. The problem statement, all variables and given/known data
    The force shown in the force-time diagram acts on a 3.4 kg object.

    p6-11.gif
    (a) Find the impulse of the force.
    (b) Find the final velocity of the mass if it is initially at rest.
    (c) Find the final velocity of the mass if it is initially moving along the x axis with a velocity of -1.8 m/s.

    2. Relevant equations

    F=ma

    Possibly kinematics

    3. The attempt at a solution
    Well I got the correct answers already:
    A) 8 (area under the line in the graph, (3*2) + (2*2/2)
    B) 2.34m/s
    C) 0.548

    The thing is I accidentally found the answers for this, I decided to start with the acceleration via:
    A=F/M
    A1=1.764
    A2=0.59

    On a whim I added these together and submitted (practice server) and it was correct, I repeated this a few times and it was always correct.

    For C) I just removed this value from my answer from B)

    What is the correct way I should have done this? I can't seem to work out the result any other way :S

    Any help is appreciated.
     
  2. jcsd
  3. Mar 26, 2015 #2
    You shouldn't use F=ma (you can, but it's not what you want). How would you be able to derive the velocity from the impulse? To think about this, how else can you write a formula for impulse?
     
  4. Mar 26, 2015 #3
    Hmmm v=u+a*t ? since I have U and T but then that falls back on acquiring A via A=f/m which isn't quite right either?

    Sorry, this particular h/w question has me stumped
     
  5. Mar 26, 2015 #4

    BvU

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    Hello Stasis, welcome to PF :smile: !

    First about a). You know about F = ma, but how did you know that the impulse of the force is the area under the graph ?

    Impulse (or momentum) is ##\vec p = m\vec v## . And acceleration is the change in ##\vec v## per unit time. ##\vec a = {d\vec v\over dt}##, to be precise. For small time steps, if the acceleration can be considered nearly constant, we write ##\Delta \vec v = \vec a \Delta t## and (in fact, even better): ##\Delta p = \vec F \Delta t##. Here we have everything in one single direction, so we can drop the vector thingy. And the area under the curve is the sum of all these little ##\Delta p = F \Delta t## contributions. In other words, $$\quad p(5) - p(0) = \int_0^5 F dt
    $$
    This is more or less the same as what you do: you take the (constant) acceleration 2/3.4 and for constant acceleration v = at so you get v = (2/3.4) * 3 which is the same as 2*3 /3.4, i.e. the area under the line / mass. And for the last two seconds you take the average acceleration (2-0)/2 = 1 to get 1/3.4 * 2 = 0.59 m/s. And (2-0)/2 * 2 is also the area under the line.


    Answer b) follows from a) if you divide by m. Since p(0) = 0 you get v(5) from this area directly.

    And answer c) follows if you substitute the proper p(0). And since m is constant, you can indeed also just subtract the 1.8 (i.e. add the -1.8 m/s).
     
  6. Mar 26, 2015 #5

    SammyS

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    Impulse is the area under the graph of Force versus time . Generally calculated (using Calculus) as ##\displaystyle\ \int_{t_1}^{t_2} \vec{F}\cdot\,dt\ ## .
    So, what you did for part (A) is absolutely correct.

    For part (B):
    Use the Impulse - Momentum Theorem, which says that the change in momentum of an object from time t1 to time t2 is equal to the Impulse of the net Force on the object from time t1 to time t2 .

    For part (C):
    What you did does give the correct answer.
    The Impulse is the same for (C) and (B), so the change in momentum is the same for both parts. Therefore, the change in velocity is the same for both parts.
    Your method gives the same change in velocity for both parts (B) and (C).
     
  7. Mar 27, 2015 #6
    Wow you're all so helpful, thank you all for the responses! I believe I understand better now :)
     
  8. Mar 27, 2015 #7

    haruspex

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    In case it is not clear from the other replies, the reason this worked is that you have calculated changes in speed, not accelerations.
    I.e., you did not use a=F/M. 1.764 is an impulse (∫F.dt) not a force (F), so the formula you really used was Δv = ∫F.dt/m. Adding the two speed changes gives the overall speed change.
     
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