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Find the work done

  1. Jun 13, 2006 #1
    The problem states:A single force acts on a 3.0 kg particle-like object in such a way that the position of the object as a function of time is given by x = 3.0t - 4.0t 2 + 1.0t 3, with x in meters and t in seconds. Find the work done on the object by the force from t = 0 to t = 3.0 s.

    I found the acceleration by double derivation of the function and then multiplied the function by 3 to give me the force and then integrated but that did not work. Any help is greatly appreciated.
     
  2. jcsd
  3. Jun 13, 2006 #2

    Hootenanny

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    You didn't need to integrate, just multiply the force you found by the distance. And allow me to take a guess that you integrated with respect to t? This is incorrect as the defintion of work is the integral of force with respect to displacement.
     
    Last edited: Jun 13, 2006
  4. Jun 13, 2006 #3
    Im sorry but for some reason im not understanding the solution you stated. I still have a variable in my force equation after deriving the accerleration and dont understand what to do next. I understand that plugging in t=3 will give me the displacement that particle traveled in that particular time interval but not what to do next.
     
  5. Jun 13, 2006 #4

    Hootenanny

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    My apologies, I forgot there was a non constant force. Now, instanataneous power is given by;

    [tex]P(t) = F(t)\cdot V(t)[/tex]

    Now,

    [tex]\text{Work} = \int P(t) \; dt[/tex]

    Can you go from here?
     
  6. Jun 13, 2006 #5

    StatusX

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    The problem may be that you need to take the absolute value of the force before integrating. Although accelerating in the negative x-direction gives a negative force, this still takes positive work, since you are doing all the work, not a conservative potential.
     
  7. Jun 13, 2006 #6
    I derived the velocity with respect to time from the function and then derived acceleration from the the velocity and multiplied the accerleration with the mass to find the force. I then multiplied the force with the velocity to find instantaneous power and integrated with respect to the displacement found by plugging in the time interval in the original position function but got a different solution. I must be doing something wrong, thanks for the help so far and I apologize if im not getting something or taking too much of your time.
     
  8. Jun 13, 2006 #7

    Hootenanny

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    Did you take the modulus of the force as Status X suggested?

    [tex]\text{Work} = \int^{3}_{0} |F(t)|\cdot V(t)[/tex]

    Since, as Status points out, although work is being done in the negative direction, work is still being done by the force, work is a scalar not a vector. Technically you are taking two vectors, (force and displacement) and turning them into a scalar by using the dot product.
     
  9. Jun 13, 2006 #8
    I dont really kow how to resolve my vector componets into something I can use the dot porudct for some reason. My force vector has one variable while my velocity vector is an exponential equation and am kinda clueless how to dot them.
     
  10. Jun 13, 2006 #9
    What do you mean your "velocity vector is an exponential equation" ?
    In this problem, we consider forces and velocities to be along only one direction and there is no need for dot products or even vectors for that matter .
    Remember velocity, V = dx/dt .
    Reread all the previous posts. The approach has been outlined clear enough .
     
  11. Jun 13, 2006 #10

    Hootenanny

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    Okay, I'll try and simplify it further for you. You have your expression for displacement with respect to time and I have given you the integral;

    [tex]\text{work} = \int^{3}_{0} |F(t)| \cdot v(t) \; dt[/tex]

    Which can be written as;

    [tex]\text{work} = \int^{3}_{0} \left| m\cdot\frac{d^{2}x}{dt^2} \right| \cdot\frac{dx}{dt} \; dt[/tex]

    Does that make it more obvious?
     
  12. Jun 13, 2006 #11
    I followed the process as stated and keep getting a wrong solution for some reason, the following are the steps I took:
    x=3t^3-4t^2+3t (dx/dt)=3t^2-8t+3 (|d^2x/d^2t|)=6t+8.
    3(6t+8)=18t+24=F V=3t^2-8t+3
    (18t+24)(3t^2-8t+3)=54t^3-120t^2-138t+72=P(t)
    Integration of P(t) yields=(54t^4/4)-(120t^2/3)-(138t^2)/(2)+72t) from 0 to 3 seconds.
     
  13. Jun 13, 2006 #12

    StatusX

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    This step isn't right. To take the absolute value, you'll have to find where the function becomes negative, and split it up around this point. For example:

    [tex] |x-3|= \left\{ \begin{array}{cc} x-3 & \mbox{ if } x>3 \\ -(x-3) & \mbox{ otherwise} \end{array}[/tex]
     
    Last edited: Jun 13, 2006
  14. Jun 13, 2006 #13
    {6t-8 if x>4/3
    so |6t-8|={-(6t-8) otherwise.
    I am not really sure how to run with this
     
  15. Jun 13, 2006 #14

    StatusX

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    Now you integrate. You'll need to split up the integral: one that goes from 0 to 4/3 and one from 4/3 to 3.
     
  16. Jun 13, 2006 #15
    Wouldnt that give me the velocity function? Am i supposed to have two integrals of 6t-8 with those bounds, add them up and multiply by the mass and original velocity function and then integrate from 0 to 3?
     
  17. Jun 14, 2006 #16

    Hootenanny

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    No, you would do the two whole integrals seperatly, thus;

    [tex]\int^{\frac{4}{3}}_{0} [ m(8-6t) ] \cdot (3t^{2} - 8t + 3) \; dt[/tex]


    [tex]\int^{3}_{\frac{4}{3}} [ m(6t - 8) ] \cdot (3t^{2} - 8t + 3) \; dt[/tex]
     
  18. Jan 16, 2007 #17
    The problem with this 'problem' is that we end up with acceleration still having a 't' in it. So what i did was just change the integration from change in distance to change in time.

    ie: dx = (3.0 - 8.0t + 3.0t^2)dt

    acceleration is
    6.0t - 8.0

    F = ma = (3.0)(6.0t - 8.0)

    I take the 3 out of the integral to make things simpler.
    Also we now have limits of integration of 0.0s to 3.0sec

    This gives us...

    W = 3.0 integral ((6.0t - 8.0)(3.0 - 8.0t + 3.0t^2)dt
    = 3.0 integral (18t^3 - 72t^2 + 82t - 24)dt
    = (3.0)(4.5t^4 - 24t^3 + 41t^2 - 24t)
    // Now we substitute 0.0sec to 3.0 sec for t and should get our answer.
     
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