JEE Mains · Physics · STD 11 - 4.2 friction
A system of two blocks of masses \(m =2 \; kg\) and \(M=8 \; kg\) is placed on a smooth table as shown in figure. The coefficient of static friction between two blocks is \(0.5\). The maximum horizontal force \(F\) that can be applied to the block of mass \(M\) so that the blocks move together will be \(\dots \; N\)
- A \(9.8\)
- B \(39.2\)
- C \(49\)
- D \(78.4\)
Answer & Solution
Correct Answer
(C) \(49\)
Step-by-step Solution
Detailed explanation
\(\left(a_{A}\right)_{\max }=0.5 g=4.9 m / s ^{2}\) For moving together \(F_{\max }=m_{T} a_{A}\) \(=10 \times 4.9\) \(=49 \; N\)
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