Question 139 of 210

A block of mass $5kg$ 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 $g=10m{s}^{-2}$]

A$25\sqrt{3}N$B$5\sqrt{3}N$C$\dfrac{5\sqrt{3}}{2}N$D$10N$

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Answer: (B) $5\sqrt{3}N$

Limiting friction force is, ${f}_{max}=\mu \left(mg\cos \theta \right)$
$=0.1\times \dfrac{50\times \sqrt{3}}{2}$
$=2.5\sqrt{3}N$
For first case:

We can write, ${F}_{1}=mg\sin \theta +{f}_{max}$
$=25+2.5\sqrt{3}$
For second case:

${F}_{2}=mg\sin \theta -{f}_{max}$
$=25-2.5\sqrt{3}$
$∴{F}_{1}-{F}_{2}=5\sqrt{3}N$