A block is moving on an inclined plane making an angle `45^@` with the horizontal and the coefficient of friction is `mu`. The force required to just push it up the inclined plane is 3 times the force required to just prevent it from sliding down. If we define `N=10mu`, then N is

Explanation

To solve the problem step by step, we will analyze the forces acting on the block on the inclined plane and derive the necessary equations.

### Step 1: Identify the forces acting on the block
On an inclined plane at an angle θ (45 degrees in this case), the forces acting on the block include:
- The gravitational force (mg) acting downward.
- The normal force (N) acting perpendicular to the surface of the incline.
- The frictional force (f) acting parallel to the incline, opposing the motion.

### Step 2: Resolve the gravitational force
The gravitational force can be resolved into two components:
- Parallel to the incline: \( mg \sin \theta \)
- Perpendicular to the incline: \( mg \cos \theta \)

Since θ = 45 degrees, we have:
- \( \sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}} \)

### Step 3: Write the equations for the forces
1. **Force required to just push the block up the incline (F1)**:
\[ F_1 = mg \sin \theta + f \]
where \( f = \mu N \) and \( N = mg \cos \theta \).

Thus, substituting for f:
\[ F_1 = mg \sin \theta + \mu (mg \cos \theta) \]

2. **Force required to just prevent the block from sliding down (F2)**:
\[ F_2 = mg \sin \theta - f \]
Again substituting for f:
\[ F_2 = mg \sin \theta - \mu (mg \cos \theta) \]

### Step 4: Set up the relationship between F1 and F2
According to the problem, \( F_1 = 3 F_2 \). Substituting the expressions for F1 and F2:
\[ mg \sin \theta + \mu (mg \cos \theta) = 3 \left( mg \sin \theta - \mu (mg \cos \theta) \right) \]

### Step 5: Simplify the equation
Cancelling \( mg \) from both sides (assuming \( mg \neq 0 \)):
\[ \sin \theta + \mu \cos \theta = 3 \left( \sin \theta - \mu \cos \theta \right) \]
Expanding the right side:
\[ \sin \theta + \mu \cos \theta = 3 \sin \theta - 3 \mu \cos \theta \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ \mu \cos \theta + 3 \mu \cos \theta = 3 \sin \theta - \sin \theta \]
This simplifies to:
\[ 4 \mu \cos \theta = 2 \sin \theta \]

### Step 7: Substitute θ = 45 degrees
Since \( \sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}} \):
\[ 4 \mu \left(\frac{1}{\sqrt{2}}\right) = 2 \left(\frac{1}{\sqrt{2}}\right) \]
Cancelling \( \frac{1}{\sqrt{2}} \) from both sides:
\[ 4 \mu = 2 \implies \mu = \frac{1}{2} \]

### Step 8: Calculate N
Given \( N = 10 \mu \):
\[ N = 10 \times \frac{1}{2} = 5 \]

### Final Answer
Thus, the value of N is \( \boxed{5} \).