A body of mass m was slowly hauled up the hill (figure) by a force 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.

Explanation

Let `vecF` makes an angle `theta` with the horizontal at any instant of time (figure). Newton's second law in projection from along the direction of the force, gives:
`F=kmgcostheta+mgsintheta` (because there is no acceleration of the body).
As `vecFuarruarrdvecr` the diffrential work done by the force `vecF`,
`dA=vecF*dvecr=Fds`, (where `ds=|dvecr|`)
`=kmgds(costheta)+mgdssintheta`
`=kmgdx+mgdy`.
Hence, `A=kmgunderset(0)overset(l)intdx+mgunderset(0)overset(h)intdy`
`=kmgl+mgh=mg(kl+h)`.