A block of mass $5 \mathrm{~kg}$ is placed on a rough inclined surface as shown in the figure.If $\vec{F}_1$ is the force required to just move the block up the inclined plane and $\vec{F}_2$ is the force required to just prevent the block from sliding down, then the value of $\left|\vec{F}_1\right|-\left|\vec{F}_2\right|$ is : [Use $g=10 \mathrm{~m} / \mathrm{s}^2$ ]
JEE MAIN 2024, Difficult
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$\mathrm{f}_{\mathrm{K}}=\mu \mathrm{mg} \cos \theta$
$=0.1 \times \frac{50 \times \sqrt{3}}{2}$
$=2.5 \sqrt{3} \mathrm{~N}$
${F}_1=\mathrm{mg} \sin \theta+\mathrm{f}_{\mathrm{K}} $
$=25+2.5 \sqrt{3}$
$\mathrm{F}_2=\mathrm{mg} \sin \theta-\mathrm{f}_{\mathrm{K}}$
$\quad=25-2.5 \sqrt{3}$
$\therefore \mathrm{F}_1-\mathrm{F}_2=5 \sqrt{3} \mathrm{~N}$
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