Question 25 of 210

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$Mg\sin \theta \sqrt{1+\mu }$

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Answer: (A) $Mg$

As the block moves with constant velocity, so force on the block should be balanced. The forces acting on the block are its weight, normal contact force and friction. The contact force is vector sum of normal contact and friction. As the forces are balanced, the contact force should be equal and opposite to its weight.
$∴$Contact force $=Mg$