A 2 kg block is pressed against a rough vertical wall by a...

Normal force exerted by the wall on the block is $N = 40 \text{ N}$.

Maximum static frictional force is f_\max = \mu_s N = 0.5 \times 40 = 20 \text{ N}.

The forces acting on the block parallel to the wall are its weight $W = mg = 2 \times 10 = 20 \text{ N}$ acting vertically downwards and the applied horizontal force $F = 15 \text{ N}$.

The net driving force on the block is $F_{net} = \sqrt{F^2 + W^2} = \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = 25 \text{ N}$.

Since F_{net} > f_\max, the block will move.

The direction of motion will be along the net driving force. Let $\theta$ be the angle it makes with the $15 \text{ N}$ force.

$\tan \theta = \dfrac{W}{F} = \dfrac{20}{15} = \dfrac{4}{3}$

\theta = 53^{}^{{}^{\circ}}

Thus, the block moves at an angle of 53^{}^{{}^{\circ}} with the $15 \text{ N}$ force.