Consider the adjacent diagram, the block is pushed up by applying a force F.

Normal reaction (N) and frictional force (f) is shown.
Given, mass 1 kg. $\mathrm{\theta } = 30°$
F = 10 N, $\mathrm{\mu }$ = 0.1 and s = distance moved by the block along the inclined plane 10 m

(a) Work done against gravity = Increase in potential energy of the block
$\begin{array}{l}=\mathrm{mg}\times \text{ Vertical distance travelled }\\ =\mathrm{mg}\times \mathrm{s}(\sin \mathrm{\theta })=\left(\mathrm{mgs}\right)\sin \mathrm{\theta }\\ =1\times 10\times 10\times \sin {30}^{\circ }=50\mathrm{J}\end{array}(\because \mathrm{g}\le 10\mathrm{m}/{\mathrm{s}}^2)$

Step 2: Find work done against friction

(b) Work done against friction
$\begin{array}{l}\mathrm{wf}=\mathrm{f}\times \mathrm{s}=\mathrm{\mu N}\times \mathrm{s}=\mathrm{\mu mgcos}\mathrm{\theta }\times \mathrm{s}\\ =0.1\times 1\times 10\times \cos {30}^{\circ }\times 10\\ =10\times 0.866=8.66\mathrm{J}\end{array}$

Step 3: Find increase in potential energy.

(c) increase in potential energy = mgh = mg (s sin $\mathrm{\theta }$)
$\begin{array}{l}=1\times 10\times 10\times \sin {30}^{\circ }\\ =100\times \frac{1}{2}=50J\end{array}$

Step 3: Find the change in kinetic energy.

(d) By work-energy theorem, we know that work done by all the forces = change in KE

$\begin{array}{l}\mathrm{W}=\mathrm{\Delta K}\\ \mathrm{\Delta k}={\mathrm{W}}_{\mathrm{g}}+{\mathrm{W}}_{\mathrm{f}}+{\mathrm{W}}_{\mathrm{f}}\\ =-\mathrm{mgh}-\mathrm{fs}+\mathrm{FS}\\ =-50-8.66+10\times 10\\ =50-8.66=41.34\mathrm{J}\end{array}$

Step 4: Find work done by the applied force.

(e) Work done by the applied force, F = FS

= (10)(10) = 100J