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$Mg \sin \theta \sqrt{1+\mu}$
JEE MAIN 2022, Medium
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A$Mg$
a
$N = Mg \cos \theta$
$f = Mg \sin \theta$
$R =\sqrt{ N ^{2}+ f ^{2}}$
$R = Mg$
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