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
JEE MAIN 2023, Difficult
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$F _1= mg \sin 45^{\circ}+ f = mg \sin 45^{\circ}+\mu N$
$F _1=\frac{ mg }{\sqrt{2}}+\mu mg \cos 45^{\circ}$
$F _1=\frac{ mg }{\sqrt{2}}(1+\mu)$
$F _2= mg \sin 45^{\circ}- f = mg \sin 45^{\circ}-\mu N$
$=\frac{ mg }{\sqrt{2}}(1-\mu)$
$F _1=2 F _2$
$\frac{ mg }{\sqrt{2}}(1+\mu)=2 \frac{ mg }{\sqrt{2}}(1-\mu)$
$1+\mu=2-2 \mu$
$\mu=1 / 3=0.33$
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