A block is moving on an inclined plane making an angle $45^\circ$ 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 $N = 10\,\mu$, then $N$ is

Step 1: Identify the given parameters.

Let the mass of the block be $m$, the acceleration due to gravity be $g$, the angle of inclination be $\theta$, and the coefficient of friction be $\mu$.

$$\theta = 45^\circ$$

Step 2: Express the force required to push the block up the incline.

The applied force $F_{\text{up}}$ must overcome both the component of gravity along the incline and the maximum static friction force acting down the incline.

$$F_{\text{up}} = mg\sin\theta + \mu mg\cos\theta$$

Step 3: Express the force required to prevent the block from sliding down.

The applied force $F_{\text{down}}$ and the friction force both act up the incline to balance the downward component of gravity.

$$F_{\text{down}} = mg\sin\theta - \mu mg\cos\theta$$

Step 4: Use the given relation to solve for the coefficient of friction.

The force to push the block up is three times the force to prevent it from sliding down.

$$F_{\text{up}} = 3F_{\text{down}}$$

Substitute the expressions for $F_{\text{up}}$ and $F_{\text{down}}$ into the relation.

$$mg\sin\theta + \mu mg\cos\theta = 3(mg\sin\theta - \mu mg\cos\theta)$$

Divide both sides by $mg$.

$$\sin\theta + \mu\cos\theta = 3\sin\theta - 3\mu\cos\theta$$

Rearrange the terms to solve for $\mu$.

$$4\mu\cos\theta = 2\sin\theta$$ $$\mu = \frac{1}{2}\tan\theta$$

Step 5: Calculate the value of $\mu$ and $N$.

Substitute $\theta = 45^\circ$ into the equation.

$$\mu = \frac{1}{2}\tan 45^\circ$$ $$\mu = 0.5$$

Calculate $N$ using the definition $N = 10\mu$.

$$N = 10 \times 0.5$$ $$N = 5$$

Answer: $N = 5$