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 land the coefficient of friction is $\mu$.

Four forces are acting on the body:

1. weight $\left(mg\right)$

2. normal reaction $\left(N\right)$

3. friction$\left(f\right)$ and

4. the applied force$\left(F\right)$

Using work-energy theorem

$W_{nd}=\Delta KE$

$W_{m_g}+W_N+W_f+W_F=0$

Here, $\Delta \mathrm{KE}=0$) because $(K_i=0=K_j)$

$$\begin{gathered} W_{mg}=-mgh \\ {W}_N=0 \end{gathered}$$

(as normal reaction is perpendicular to displacement at all points)

Wy can be calculated as under

$$\begin{gathered} f=\mu mg\cos \theta \\ {\left(d{W}_{AB}\right)}_f=-fds \\ =-\left(\mu mg\cos \theta \right)ds \\ =-\mu mg\left(dl\right) \end{gathered}$$

f =$-\mu mg\Sigma dl=-\mu mgl$

Substituting these values in Eq. (i), we get

$W_F=mgh+\mu mgl$