PhysicsEasy68×since 2002Q7569A block of mass M slides down on a rough inclined plane with constant velocity. The angle made by the incline plane with horizontal is θ\thetaθ. The magnitude of the contact force will be :AMgBMgcosθ\mathrm{Mg} \cos \thetaMgcosθCMgsinθ+Mgcosθ\sqrt{\mathrm{Mg} \sin \theta+\mathrm{Mg} \cos \theta}Mgsinθ+MgcosθDMgsinθ1+μ\operatorname{Mg} \sin \theta \sqrt{1+\mu}Mgsinθ1+μCheck answerSkip