Inclined Plane with Friction

At drawing an FBD with $x$ axis parallel to the surface where is falling the object we can realize that there is a normal force, gravity force and friction force; with $N=cos(\theta)mg$ and therefore $f=\mu \cdot cos(\theta)mg$ . We are given all these values, except $\theta$ , but we can see that the inclined plane is a right triangle, so the hypotenuse equals $10[m]$ and then $cos(\theta)=\frac{8}{10}=0,8$ , so replacing the values and using the fact that friction force opposes to the movement, we have: $\vec{f}=-0,6\cdot 0,8 \cdot 50[N]=-24$ $\Rightarrow W_{f}=\vec{f}\cdot \vec{r}=-24 \cdot 10[J]=-240[J]$