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Forces problem

  1. Feb 10, 2008 #1
    1. The problem statement, all variables and given/known data
    The force exerted on a 10kg mass is given by F=10 + 2t where the units are SI. If the mass starts from rest, its velocity after 2s is:
    A) 14 m/s
    B) 2 m/s
    C) 2.4 m/s
    D) .2 m/s
    E) .24 KM/S
    2. Relevant equations
    F=ma
    thats it?

    3. The attempt at a solution
    At 2 secs the force would be 14 N so..well, im not sure on what to do. i could probably figure it out if i knew what i was supposed to do though. Im not sure how a varying force could be ...i have no idea what im talking about :(. Any help would be greatly appriciated.
     
  2. jcsd
  3. Feb 10, 2008 #2
    Well, you know the mass, so you can find the acceleration of the object. You can probably use a kinematics equation that will tell you it's velocity after 2 seconds. (remember it starts at rest.
     
  4. Feb 10, 2008 #3
    if you mean like v(f)=v(i) + at, then i cant use that because i dont have the acceleration or the final velocity, the force that acts on it is external so...am i confusing my self here?
     
  5. Feb 10, 2008 #4

    rock.freak667

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    Homework Helper

    Well you found the force at t=2 which is F=14...and F=ma...you can find acceleration now...
     
  6. Feb 10, 2008 #5
    OK, the probelm with this question is the variable force.

    The way I see this there are two ways to solve it. One is easy, one is harder.

    The easy way is to notice that the force increases linearly with time, therefore you can find the average acceleration and simply use that in the F = ma equation. As you are looking at a period of 2 seconds, and the average occurs mid-way between 0 seconds and 2 seconds, I'll let you take it from here.

    The harder way is to integrate F = 10 + 2t with respect to t. For which you get Ft = 10t + t^2, since Ft (Force * time) is an impulse, you can then use Ft = mv to come to your answer. 10t + t^2 = m * v, and solve for v where m = 10 and t = 2.
     
  7. Feb 10, 2008 #6
    you guys are awesome, thanks a ton :D
     
  8. Feb 11, 2008 #7
    Ah, I did my integration a bit hurridly, what I should have said was:
    [tex]Ft = 10t + t^{2} + c[/tex]

    Then when t = 0:
    [tex]0 = 0 + 0 + c[/tex]
    Therefore:
    [tex]c = 0[/tex]
    Therefore:
    [tex]Ft = 10t + t^{2}[/tex]
     
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