Question

A body of mass $m$ was slowly hauled up the hill (figure shown above) by a force $\overrightarrow{F}$ which at each point was directed along a tangent to the trajectory. Find the work performed by this force, if the height of the hill is $h$, the length of its base $l$, and the coefficient of friction $k$.

Answer 1

Let $\overrightarrow{F}$ makes an angle $\theta$ with the horizontal at any instant of time (figure shown below). Newton's second law in projection form along the direction of the force, gives
$F=kmg\cos \theta +mg\sin \theta$ (because there is no acceleration of the body.)
As $\overrightarrow{F}\upuparrows d\overrightarrow{r}$ the differential work done by the force $\overrightarrow{F}$,
$dA=\overrightarrow{F}\cdot d\overrightarrow{r}=Fds$, (where $ds=|d\overrightarrow{r}|$)
$=kmgds\left(\cos \theta \right)+mgds\sin \theta$
$=kmgdx+mgdy$.
Hence, $A=kmg{\int }_0^{l}dx+mg{\int }_0^{h}dy$
$=kmgl+mgh=mg(kl+h)$