Consider a moving particle m under the influence of gravity. What's the work done by gravity in moving m from A to B?

$$ \int_A^B \vec{F}\cdot \vec{dS}= \int_{rA}^{rB} \vec{F}\,dr $$

I din't understand this change of variables. Why the work can be calculated using dr instead of dS without considering the angle between dS and dr?

Thank you!

  • $\begingroup$ What are $S$ and $r$? $\endgroup$ – user7530 Apr 21 '17 at 16:08
  • 2
    $\begingroup$ This is likely a central force problem where the gravitational force is in the r direction $\endgroup$ – Paul Apr 21 '17 at 16:11

Imagine that the path connecting $A$ and $B$ is a curve $c$. At any point ${\bf r}$ the direction tangent to the curve $c$ can be written as the linear combination of the radial unitary vector $\hat{\bf r}$, and a perpendicular vector $\hat{\bf r}_\perp$ (in three dimensions, this is actually two vectors).

$$ d{\bf S} = dr\hat{\bf r} + dq \hat{\bf r}_\perp $$

Now, the gravitational force is radial so ${\bf F} = F{\bf \hat{r}}$, this means that

$$ {\bf F}\cdot d{\bf S} = (F{\bf \hat{r}})\cdot (dr\hat{\bf r} + dq \hat{\bf r}_\perp) = Fdr $$

In other words

$$ \int_A^B {\bf F}\cdot d{\bf S} = \int_{r_A}^{r^B}F dr $$


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