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How does the kinematic energy appear in the work integral?

  1. Jan 21, 2016 #1
    1. The problem statement, all variables and given/known data

    We call Kinetic energy Ec=1/2 mv^2

    its deferral into the work integral gives:

    \begin{align*}
    w_{12}&=\int\limits_1^2\delta W\\
    & = \int\limits_1^2 \vec F \wedge d\vec r\\
    &\underbrace{=\int\limits_1^2dE_c}_{why?}
    \end{align*}

    But Why does the kinematic energy appears in the last equation?


    2. Relevant equations

    According to Newton's second law,

    $$\vec{F}=\frac{d\vec{P}}{dt}$$


    3. The attempt at a solution

    Multiplicating it by $$d\vec{r}$$

    \begin{align*}
    \vec F \wedge d\vec r &= \frac{d\vec P}{dt} \wedge d\vec r\\
    &= d\vec P \wedge \frac{d\vec r}{dt}\\
    &= d\vec P \wedge \vec v\\
    &= d(m\vec v) \wedge \vec v\\
    \end{align*}



    I'm so sorry but I don't know from there..!
     
  2. jcsd
  3. Jan 21, 2016 #2

    cnh1995

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    Work done on the body results in change in it's kinetic energy.
     
  4. Jan 21, 2016 #3

    PeroK

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    If it's any consolation, none of those integrals makes mathematical sense to me. Instead:

    ##\vec{F}.d\vec{r} = m\frac{d\vec{v}}{dt}.\frac{d\vec{r}}{dt}dt = m\frac{d\vec{v}}{dt}.\vec{v}dt = \frac{1}{2}m\frac{d}{dt}(\vec{v}.\vec{v})dt##

    And, then if you do your line integral with ##t## as your parameter it comes out.
     
  5. Jan 21, 2016 #4

    PeroK

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    ##\int_C \vec{F}.d\vec{r} = \int_{t_1}^{t_2}\frac{1}{2}m \frac{d}{dt}(v^2)dt##

    That's the mathematical format that I'm used to and makes sense to me. A line integral depends on the curve C not just its end- points.
     
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