JEE Mains · Physics · STD 11 - Friction
- JEE Mains 2024
- Hard
- MCQ
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 \(g=10 \mathrm{~m} / \mathrm{s}^2\) ]
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(5 \sqrt{3} \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{f}_{\mathrm{K}}=\mu \mathrm{mg} \cos \theta\)
\(=0.1 \times \frac{50 \times \sqrt{3}}{2}\)
\(=2.5 \sqrt{3} \mathrm{~N}\)
\({F}_1=\mathrm{mg} \sin \theta+\mathrm{f}_{\mathrm{K}} \)
\(=25+2.5 \sqrt{3}\)
\(\mathrm{F}_2=\mathrm{mg} \sin \theta-\mathrm{f}_{\mathrm{K}}\)
\(\quad=25-2.5 \sqrt{3}\)
\(\therefore \mathrm{F}_1-\mathrm{F}_2=5 \sqrt{3} \mathrm{~N}\)
\(=0.1 \times \frac{50 \times \sqrt{3}}{2}\)
\(=2.5 \sqrt{3} \mathrm{~N}\)
\({F}_1=\mathrm{mg} \sin \theta+\mathrm{f}_{\mathrm{K}} \)
\(=25+2.5 \sqrt{3}\)
\(\mathrm{F}_2=\mathrm{mg} \sin \theta-\mathrm{f}_{\mathrm{K}}\)
\(\quad=25-2.5 \sqrt{3}\)
\(\therefore \mathrm{F}_1-\mathrm{F}_2=5 \sqrt{3} \mathrm{~N}\)
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