JEE Mains · Physics · STD 11 - Friction
- JEE Mains 2023
- Hard
- MCQ
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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Answer & Solution
Correct Answer
(A) \(0.33\)
Step-by-step Solution
Detailed explanation
\(F _1= mg \sin 45^{\circ}+ f = mg \sin 45^{\circ}+\mu N\)
\(F _1=\frac{ mg }{\sqrt{2}}+\mu mg \cos 45^{\circ}\)
\(F _1=\frac{ mg }{\sqrt{2}}(1+\mu)\)
\(F _2= mg \sin 45^{\circ}- f = mg \sin 45^{\circ}-\mu N\)
\(=\frac{ mg }{\sqrt{2}}(1-\mu)\)
\(F _1=2 F _2\)
\(\frac{ mg }{\sqrt{2}}(1+\mu)=2 \frac{ mg }{\sqrt{2}}(1-\mu)\)
\(1+\mu=2-2 \mu\)
\(\mu=1 / 3=0.33\)
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