A block of mass M slides down a rough incline | vedclass

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(A) Since the block slides down with constant velocity,the net force acting on it is zero.The forces acting on the block are the gravitational force $

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$Mg$

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(A) Since the block slides down with constant velocity,the net force acting on it is zero.The forces acting on the block are the gravitational force $(Mg)$,the normal force $(N)$,and the kinetic friction force $(f)$.Resolving the gravitational force into components,we have:$N = Mg \cos 	heta$ (perpendicular to the plane)$f = Mg \sin 	heta$ (parallel to the plane)The contact force $(R)$ is the resultant of the normal force $(N)$ and the friction force $(f)$:$R = \sqrt{N^2 + f^2}$Substituting the values of $N$ and $f$:$R = \sqrt{(Mg \cos 	heta)^2 + (Mg \sin 	heta)^2}$$R = \sqrt{M^2g^2(\cos^2 	heta + \sin^2 	heta)}$Since $\sin^2 	heta + \cos^2 	heta = 1$,we get:$R = \sqrt{M^2g^2(1)}$$R = Mg$

$A$ block of mass $M$ slides down a rough inclined plane with constant velocity. The angle made by the inclined plane with the horizontal is $\theta$. The magnitude of the contact force will be:

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