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 horizontal force that should be applied on the block to keep it adhere to the wall will be ____N.

Answer

To solve the problem, we need to determine the magnitude of the horizontal force (F) that should be applied on a wooden block of mass 0.5 kg to keep it adhered to a vertical rough wall, given that the coefficient of static friction (μ) between the block and the wall is 0.2.

### Step-by-Step Solution:

1. **Identify the forces acting on the block:**
- The weight of the block (W) acting downwards:
\[ W = mg \]
where \( m = 0.5 \, \text{kg} \) and \( g = 10 \, \text{m/s}^2 \).
- The normal force (N) acting perpendicular to the wall, which is equal to the applied horizontal force (F).
- The frictional force (f) acting upwards, which opposes the weight of the block.

2. **Calculate the weight of the block:**
\[ W = mg = 0.5 \, \text{kg} \times 10 \, \text{m/s}^2 = 5 \, \text{N} \]

3. **Determine the frictional force:**
The maximum static frictional force can be calculated using the formula:
\[ f = \mu N \]
where \( \mu = 0.2 \).

4. **Set up the equilibrium condition:**
For the block to remain at rest, the upward frictional force must balance the downward weight:
\[ f = W \]
Substituting the expression for friction:
\[ \mu N = mg \]

5. **Substituting the normal force:**
Since the normal force \( N \) is equal to the applied horizontal force \( F \):
\[ \mu F = mg \]

6. **Rearranging the equation to find F:**
\[ F = \frac{mg}{\mu} \]

7. **Substituting the known values:**
\[ F = \frac{5 \, \text{N}}{0.2} = 25 \, \text{N} \]

### Final Answer:
The magnitude of the horizontal force that should be applied on the block to keep it adhered to the wall is **25 N**.