Question

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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Answer 1

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") = DeltaKE`
or `W_mg + W_N + W_f + W_f = `…….(i)
Here, `DeltaKE = 0`, because `K_(i) = 0 K_(f)`
`W_(mg) = - mgh`
`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`
`:. (dW_(AB))_(f) = - fds`
`= -(mu mg cos theta)ds`
`= -mu mg(dl)` , (as `ds cos theta = dl`)
`:. f = - mu mg sum dl = - mu mgl`
Substituting these values in Eq. `(i)`, we get
`W_(F) = mgh + mu mgl`