JEE Mains · Physics · STD 11 - 4.2 friction
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 \(\left.g =10\,ms ^{-2}\right]\)
- A \(33.3\)
- B \(25.2\)
- C \(20\)
- D \(35.7\)
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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