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, we need to analyze the forces acting on the block on the inclined plane. Let's break it down step by step.

### Step 1: Identify the forces acting on the block
1. **Weight (W)**: The weight of the block acts vertically downward and is given by \( W = mg \), where \( m \) is the mass of the block and \( g \) is the acceleration due to gravity.
2. **Normal Force (N)**: The normal force acts perpendicular to the surface of the incline.
3. **Frictional Force (F_f)**: The frictional force acts parallel to the surface of the incline and opposes the motion. It is given by \( F_f = \mu N \), where \( \mu \) is the coefficient of friction.

### Step 2: Resolve the weight into components
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