JEE Mains · Physics · STD 11 - 4.2 friction
A block of mass \(M\) slides down on a rough inclined plane with constant velocity. The angle made by the incline plane with horizontal is \(\theta\). The magnitude of the contact force will be.
- A \(Mg\)
- B \(Mg \cos \theta\)
- C \(\sqrt{ Mg \sin \theta+ Mg \cos \theta}\)
- D \(Mg \sin \theta \sqrt{1+\mu}\)
Answer & Solution
Correct Answer
(A) \(Mg\)
Step-by-step Solution
Detailed explanation
\(N = Mg \cos \theta\) \(f = Mg \sin \theta\) \(R =\sqrt{ N ^{2}+ f ^{2}}\) \(R = Mg\)
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