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 use the given information to find the value of \( N \).

### Step 1: Identify the Forces Acting on the Block
On an inclined plane at an angle \( \theta = 45^\circ \):
- The weight of the block \( mg \) can be resolved into two components:
- Perpendicular to the incline: \( mg \cos \theta \)
- Parallel to the incline: \( mg \sin \theta \)
- The normal force \( N \) acts perpendicular to the incline.
- The frictional force \( f \) acts parallel to the incline, opposing the motion.

### Step 2: Write the Equations for Forces
1. **Force to Prevent Sliding Down (F1)**:
To prevent the block from sliding down, the frictional force must balance the component of weight acting down the incline:
\[ F_1 = mg \sin \theta - f \]
where \( f = \mu N \).

2. **Force to Push Up the Incline (F2)**:
To push the block up the incline, the applied force must overcome both the gravitational component and the frictional force:
\[ F_2 = mg \sin \theta + f \]

### Step 3: Relate the Two Forces
According to the problem, the force required to push the block up the incline \( F_2 \) is three times the force required to prevent it from sliding down \( F_1 \):
\[ F_2 = 3 F_1 \]

### Step 4: Substitute the Forces
Substituting the expressions for \( F_1 \) and \( F_2 \):
\[ mg \sin \theta + \mu N = 3(mg \sin \theta - \mu N) \]

### Step 5: Simplify the Equation
Expanding and rearranging the equation:
\[ mg \sin \theta + \mu N = 3mg \sin \theta - 3\mu N \]
\[ mg \sin \theta + 4\mu N = 3mg \sin \theta \]
\[ 4\mu N = 2mg \sin \theta \]
\[ \mu N = \frac{1}{2} mg \sin \theta \]

### Step 6: Substitute \( \theta = 45^\circ \)
Since \( \sin 45^\circ = \frac{1}{\sqrt{2}} \):
\[ \mu N = \frac{1}{2} mg \cdot \frac{1}{\sqrt{2}} = \frac{mg}{2\sqrt{2}} \]

### Step 7: Given Relation for N
We are given that \( N = 10\mu \). Substituting this into the equation:
\[ \mu (10\mu) = \frac{mg}{2\sqrt{2}} \]
\[ 10\mu^2 = \frac{mg}{2\sqrt{2}} \]

### Step 8: Solve for \( N \)
From the above equation, we can express \( N \):
\[ N = 10\mu \]
To find \( N \), we need to express \( \mu \):
\[ \mu = \frac{mg}{20\sqrt{2}} \]
Substituting back into \( N \):
\[ N = 10 \left(\frac{mg}{20\sqrt{2}}\right) = \frac{mg}{2\sqrt{2}} \]

### Step 9: Final Calculation
We can express \( N \) in terms of \( g \) (taking \( g \approx 10 \, \text{m/s}^2 \)):
\[ N = 10 \mu = 10 \left(\frac{10}{20\sqrt{2}}\right) = \frac{100}{20\sqrt{2}} = \frac{5}{\sqrt{2}} \approx 3.54 \, \text{N} \]

### Conclusion
Thus, the value of \( N \) is approximately \( 5 \, \text{N} \).