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Given Force is 4xDisplacement, find work done

  1. Dec 28, 2014 #1
    1. The problem statement, all variables and given/known data
    A force ##F=4x## is applied on an object. What is work done to move it from ##x_1=2m## to ##x_2=4m##?

    A)12 J
    B)24 J
    C)32 J
    D)48 J
    2. Relevant equations


    3. The attempt at a solution
    The displacement, ##x=4-2=2m##
    Force, ##F=4x=4 \times 2=8N##
    Work ##W=\vec F \cdot \vec x=8 \times 2 = 16 J##
    But it is not listed in options. Did I miss anything? I guess it because Force was given as 4 times x, which will result in meters than Newtons.
     
  2. jcsd
  3. Dec 28, 2014 #2

    ehild

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    The force is not constant, so you have to calculate the work as ##W=\int_2^4F(x)dx##

    F=4x means that 4 has the proper unit, N/m.
     
  4. Dec 28, 2014 #3
    Horrible! Calculus is not in the portion. anyway I know some basics of it and figured out the answer as 32-8=24 J. Thank you very much.
     
  5. Dec 29, 2014 #4

    ehild

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    If you haven't studied Calculus yet, plot the function F(x) and find the area under the plot between the limits x=2 and x=4. It is a trapezoid now.
     
  6. Dec 29, 2014 #5
    I have one doubt. What is the correct definition of work?
    My books say ##W=\vec F \cdot \vec x##
    But ##W=\int F(x) dx## seems more reasonable.

    Also what if force is given as a function of time.
     
  7. Dec 29, 2014 #6

    ehild

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    ##W= \vec F \cdot \vec x ## if the force is constant and ##\vec x## is the displacement.
    If the force changes with the position x, the elementary work done during a small displacement ##\vec {dx} ## is ##dW=\vec F \cdot \vec {dx}##, the scalar product of the force with the small displacement.More accurately
    To get the whole work you have to integrate for the whole path..

    If the force depends on time, you change for the variable t:
    ##dW=\vec F \cdot \vec {dx}= \vec F \cdot (\vec {dx}/dt )dt=\vec F \cdot \vec {v} dt= ##
     
    Last edited: Dec 29, 2014
  8. Dec 29, 2014 #7
    Amazing!, how to work on the last equation? Is this correct?
    $$dW=\vec F \cdot \vec v dt$$
    $$\frac{dW}{dt}=\vec F \cdot \vec v$$[Can I just divide the equation by ##dt## like this one?\
    $$\text{Power}=\vec F \cdot \vec v$$
    Is this also right?
    [tex]
    \begin{align*}\\\frac{dW}{dt}&=\vec F \cdot \vec v\\\\
    \frac{dW}{dt}&=\frac{m\cdot \vec{dv} }{dt}\cdot \vec v\\\\
    dW&=m\cdot dv^2
    \end{align*}\\
    [/tex]
     
  9. Dec 29, 2014 #8
    Actually you have to take average of displacement means x=X1+X2/2 than you find the answer it will be 24.
     
  10. Dec 29, 2014 #9

    ehild

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    Yes, it is right. The power is the scalar product of force with velocity.. It is a very useful equation!

    It is not right, as
    ## \vec v \cdot \frac{d \vec v}{dt} = 0.5 \frac {d(\vec v)^2} {dt}##,

    So ##dW=0.5 m d(\vec v)^2= d KE##

    This is the Work-Energy Theorem. The change of kinetic energy is equal to the work done by the force (resultant of all external forces) . .
     
  11. Dec 29, 2014 #10
    I think x1, x2 are position rather than displacement.
    Displacement is the change in position, isn't it?
     
  12. Dec 29, 2014 #11

    ehild

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    x1 and x2 are positions. The displacement is x2-x1.
     
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