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Calculating Work

  1. Jan 8, 2006 #1
    The froce of friction on a body (moving along a surface at constant speed) can be approximated by [itex] \vec{F} = -F \hat{v}[/itex], where F is a positive constant and [itex] \hat{v} = \frac{\vec{v}}{|\vec{v}|} [/itex] is the unit vector in teh direction of [itex] \vec{v} = \dot{\vec{r}} [/itex] (the velocity of the body)

    Prove that hte work done in moving the body from r to r+dr (along a given path) ie dW = F . dr= -F ds in this case where
    [itex] ds^2 = d\vec{r}^2 = dx^2 + dy^2 + dz^2 [/itex] where ds iteh element of arc length along the path of the body


    Hint: [tex] \frac{\frac{d\vec{r}}{dt}}{|\frac{d\vec{r}}{dt}|} = \frac{d\vec{r}}{|d\vec{r}|} [/tex] and [tex] |d\vec{r}| = \sqrt{d\vec{r} \bullet d\vec{r}} = \sqrt{d\vec{r}^2} = ds [/tex]

    ok since v hat can be written like the hint and the denominator can be written like ds so
    [tex] \vec{F} \bullet d\vec{r} = -F \frac{d\vec{r}}{ds} \bullet d\vec{r} [/tex]
    since dr dot dr is ismply dr^2 which is ds^2 we get what we want.
    Is this fine?

    If the body is moved along a level floor (in which we take to be the xy plane) from the point (0,0) to (1,1) along two different paths
    (i) The straight line through (0,0) and (1,1)
    (ii) along teh parabola y = x^2
    Calculate the work done in teh two cases and compare the results


    Im wondering if they mean the work done by the force of friction of the work done by the person pushing the box

    If it was the work done by the friction could we even calculate it without any time given?

    Please help! Thanks in advance!
     
  2. jcsd
  3. Jan 9, 2006 #2
    i asked my prof and he says that the work done should be the work done by the friction

    so in this case for the straight line y=x, since we derivaed the dW = -F ds
    but what is ds?
    is it something like this?
    [tex] ds = \sqrt{dx^2 + dy^2} [/tex]

    in cartesian coordiantes what would be the limits of integration... since x and y would be inmtergrated separately?

    so then would it be like this?
    [tex] W = \int_{y=0}^{y=1} \int_{x=0}^{x=1} \sqrt{dx^2 + dy^2} [/tex]
    but htis doesnt make sense because the differe\ntials themselves cant be under roots?
    Please help!
    Thank you!
     
  4. Jan 11, 2006 #3
    can anyone help... offer some guidance as to what to do here?

    your help and input is always greatly appreciated!
     
  5. Jan 11, 2006 #4
    a friend of mine said that the answer for the part where the block is moved along the straight path form (0,0) to (1,1)
    [tex] W = \int_{0}^{1} \int_{0}^{1} F (1,1) \bullet (1,1) dx dy [/tex] but im not quite sure about why he did the dot product?

    for the y=x^2 path he suggested that one does this
    [tex] W = \int_{0}^{1} \int_{y=0}^{y=x^2} F (1,1) \bullet (1,1) dx dy [/tex]

    same thing here, how did he get the dot product?
     
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