JEE Mains · Physics · STD 11 - Friction
- JEE Mains 2023
- Medium
- MCQ
As shown in the figure a block of mass \(10\,kg\) lying on a horizontal surface is pulled by a force \(F\) acting at an angle \(30^{\circ}\), with horizontal. For \(\mu_{ s }=0.25\), the block will just start to move for the value of \(F..........\,N\) : \(\left[\right.\) Given \(\left.g =10\,ms ^{-2}\right]\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(25.2\)
Step-by-step Solution
Detailed explanation
\(N = Mg - F Sin 30^{\circ}\)
\(= mg -\frac{ F }{2}=100-\frac{ F }{2}=\frac{200- F }{2}\)
\(F Cos 30^{\circ}=\mu N\)
\(\sqrt{3} \frac{ F }{2}=0.25 \times\left(\frac{200- F }{2}\right)\)
\(4 \sqrt{3} F =200- F\)
\(F =\frac{200}{4 \sqrt{3}+1}=25.22\)
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