The coefficient of static friction between a wooden block of mass 0.5 kg and a vertical rough wall is 0.2. The magnitude of the horizontal force that should be applied on the block to keep it adhere to the wall will be ______ N. $$g = 10$$ m s$$^{-2}$$

We have a wooden block of mass $$m = 0.5$$ kg pressed against a vertical rough wall by a horizontal force $$F$$. The coefficient of static friction is $$\mu_s = 0.2$$.

When the block is pushed horizontally against the wall, the normal reaction from the wall equals the applied force. So we have $$N = F$$.

The frictional force acts upward along the wall to prevent the block from sliding down. The maximum static friction is $$f = \mu_s N = \mu_s F$$.

For the block to remain stationary on the wall, the friction must balance the weight of the block. So $$f \geq mg$$, which gives us $$\mu_s F \geq mg$$.

Substituting the values, we get $$0.2 \times F \geq 0.5 \times 10$$.

$$0.2F \geq 5$$

$$F \geq \frac{5}{0.2} = 25 \text{ N}$$

The minimum horizontal force required is $$F = 25$$ N.

So, the answer is $$25$$.