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Dynamic problem

  1. May 25, 2007 #1
    1. The problem statement, all variables and given/known data
    an object with m=100kg is moving on a straight line. for t=0 its speed is v0=20 m/s. There's a force which is opposing to the movement F=C*v^2 ( C=1kg/m and v is the speed (v(t)=...) ). Find the time t1 in which the speed v=v0/2.


    2. Relevant equations
    F=m*a (?)
    v(t)=a*t+v0


    3. The attempt at a solution
    v1=a*t1+v0 --> v0/2=a*t1+v0 --> t1=(-v0/2)*1/a

    assuming a (accelation) constant
    m*a = F = C*v1^2= C*(v0/2)^2
    a= [C*(v0/2)^2]/m
    Putting a in
    t1= (-v0/2)/a should be getting the answer

    Unfortunately the answer I get is not correct (should be 5second)
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 25, 2007 #2

    cepheid

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    The acceleration is not constant because the force is not constant. Think about it. The drag force is proportional to the square of the velocity.This means that as the object slows down (due to the drag force), the drag force itself decreases. So not only does it decelerate, but it decelerates at an ever decreasing rate. You need a new way to approach this problem. Using the dynamical equations derived for the special case of motion under a constant force doesn't cut it here.
     
  4. May 25, 2007 #3
    then how to solve it?
     
  5. May 25, 2007 #4
    Big hint: Think Newton's Second Law and think calculus....
     
  6. May 25, 2007 #5
    F=m*a=C*v^2
    so a=(C*v^2)/m

    a=dv/dt ==> v=integral ( (C*v^2)/m )= C*v^2/m * t

    so then

    t= (v*m)/(C*v^2)=m/(C*v)

    I then say
    t0=0
    t1 -t0=t1
    t0=m/(C*v0)
    t1= 2m/(C*v0)
    t1= t1-t0= m/(C*v0)=5 seconds
     
  7. May 25, 2007 #6
    Umm..not quite.
    v is actually dependent on t, and so you can't treat it as a constant.
    You should write
    [tex]\frac{dv}{dt}=Cv^2/m[/tex]

    [tex]\frac{dv}{v^2}=Cdt/m[/tex]
    And integrate on both sides.

    And don't forget the constant of integration whose value can be determined with the given data.
     
  8. May 25, 2007 #7
    [tex]\int{\frac{1}{v^2}}dv[/tex]? How to do it? I dunno
    Maybe [tex]-\frac{1}{v}[/tex]?

    Then what about [tex]\int \frac{C}{m}dt= \frac{C}{m} \cdot t+ c_0[/tex]?
     
    Last edited: May 25, 2007
  9. May 25, 2007 #8
    You are perfectly correct, why did the doubts arise?
    Anyway can you proceed now?
     
  10. May 25, 2007 #9
    So is it ....

    [tex]-\frac{1}{v}+v_0=\frac{C}{m} \cdot t [/tex]

    [tex] t=\frac{m}{C} \cdot ( -\frac{1}{v}+v_0)[/tex] ?

    Then I substitute [tex]v=\frac{v_0}{2}[/tex] in the last expression and I should get the answer, right?
    result should be 5 seconds.
    Let's see...
    Well...it's not right
     
    Last edited: May 25, 2007
  11. May 25, 2007 #10
    I notice that in the question, it is mentioned that the mass is decelarated. This means you should assign a negative value to acceleration.

    Not quite, v_0 refers to velocity at time t=0.

    Integrating, you should get
    -1/v + K(constant of integration) = -Ct/m

    Can you find K?
     
  12. May 25, 2007 #11
    Thought it was K=v0; wouldn't know How can I get it ?
     
  13. May 25, 2007 #12
    At t=0, v=20. For LHS=RHS, K= ....
     
  14. May 25, 2007 #13
    What's LHS=....? :eek:
    Never heard
    Is it K=20? but that's what i said previously: K=v0=20
     
  15. May 25, 2007 #14
    LHS - Left hand side RHS - Right hand side (for the equation -1/v + K(constant of integration) = -Ct/m) ) :smile:
    What I'm saying is put t=0 and v=20 in this equation and find K.
     
  16. May 25, 2007 #15

    cepheid

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    LHS - left-hand side (of the equation), that's all
     
  17. May 26, 2007 #16
    so k=1/20
    [tex]-\frac{1}{v} +\frac{1}{20}=-\frac{C}{m} \cdot t[/tex]

    Is it right?
    Now I substitute t=t1 and v=v0/2 in the expression?

    [tex]-\frac{1}{v} +\frac{1}{20}=-\frac{C}{m} \cdot t1[/tex]

    [tex]t1=(-\frac{m}{C}) \cdot (-\frac{1}{v} +\frac{1}{20}) =[/tex]

    [tex]=(-\frac{m}{C}) \cdot (-\frac{2}{v_0}+\frac{1}{20})=[/tex]

    [tex]=(-\frac{m}{C}) \cdot (\frac{-2 \cdot 20 + v_0}{20 \cdot v_0})=[/tex]

    [tex]=(-\frac{m}{C}) \cdot (\frac{-40+v_0}{20 \cdot v_0})=[/tex]

    [tex]=(-100) \cdot (\frac{-40+20}{20 \cdot 20})=[/tex]

    [tex]=(-100) \cdot (\frac{-20}{400})= 5 \mbox{ seconds }[/tex]

    OK! GREAT! Thank you! :blushing: :rofl:
     
    Last edited: May 26, 2007
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