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(A) Let $m$ be the mass of the block and $\mu$ be the coefficient of friction.Case $1$: Force $F_1$ required to just push the block up the incline.The

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$0.33$

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(A) Let $m$ be the mass of the block and $\mu$ be the coefficient of friction.Case $1$: Force $F_1$ required to just push the block up the incline.The forces acting along the incline are $F_1$ (upwards),$mg \sin 45^{\circ}$ (downwards),and frictional force $f = \mu N = \mu mg \cos 45^{\circ}$ (downwards).$F_1 = mg \sin 45^{\circ} + \mu mg \cos 45^{\circ} = \frac{mg}{\sqrt{2}}(1 + \mu)$Case $2$: Force $F_2$ required to just prevent the block from sliding down.The forces acting along the incline are $F_2$ (upwards),$mg \sin 45^{\circ}$ (downwards),and frictional force $f = \mu N = \mu mg \cos 45^{\circ}$ (upwards).$F_2 = mg \sin 45^{\circ} - \mu mg \cos 45^{\circ} = \frac{mg}{\sqrt{2}}(1 - \mu)$Given that $F_1 = 2 F_2$,we have:$\frac{mg}{\sqrt{2}}(1 + \mu) = 2 \left[ \frac{mg}{\sqrt{2}}(1 - \mu) \right]$$1 + \mu = 2(1 - \mu)$$1 + \mu = 2 - 2\mu$$3\mu = 1$$\mu = \frac{1}{3} \approx 0.33$

Consider a block kept on an inclined plane (inclined at $45^{\circ}$) as shown in the figure. If the force required to just push it up the incline is $2$ times the force required to just prevent it from sliding down,the coefficient of friction between the block and inclined plane $(\mu)$ is equal to

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