A body of mass m was slowly hauled up the hill as shown in the fig. 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 is I and the coefficient ot friction is `mu`.
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Explanation

Four forces are acting on the body:
1. weight (mg) `" "` 2. normal reaction (N)
3. friction (f) and `" " ` 4. the applied force (F)
Using work-energy theorem
`W_("net") = Delta KE`
or `" " W_("mg") + W_(N) + W_(f) + W_(F) = 0`
here, `Delta` KE = 0, because `K_(i) = 0 = K_(f)`
`W_("mg") = 0 "mgh" rArr W_(N) = 0 `
(as normal reaction is perpendicular to displacement at all points)
`W_(f) `can be calculated as under
`f = mu " mg " cos theta `
`thetefore " " ("d"W_(AB))_(f) = -f` ds
= - `(mu " mg cos " theta) "ds " = - mu "mg"` (dl)
(as ds cos `theta` = dl)
`therefore " " f = - mu "mg" sum "dl " = - mu` mgl
Substituting these values is Eq. (i) , we get
`W_(f) = " mgh " + mu"mgl"`
Note: Here again, if we want to solve this problem without using work-energy theorem we will first find magnitude of applied force `vec(F)` at different locations and then integrate dW ( = `vec(F).d vec(r )`) with proper limits