A block of mass m is stationary on a rough pl | vedclass

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(D) In the frame of the inclined plane,the block experiences a pseudo force $ma$ downwards. The effective acceleration due to gravity is $(g+a)$ downw

Vedclass · Accepted Answer

resultant of $m(g+a) \cos 	heta$ normal to the plane and $\mu m(g+a) \cos 	heta$ along the plane

Vedclass · Answer

(D) In the frame of the inclined plane,the block experiences a pseudo force $ma$ downwards. The effective acceleration due to gravity is $(g+a)$ downwards.Resolving the effective weight $m(g+a)$ into components perpendicular and parallel to the inclined plane:$1$. The component perpendicular to the plane is $N = m(g+a) \cos 	heta$. This is the normal force exerted by the plane on the block.$2$. The component parallel to the plane is $m(g+a) \sin 	heta$. This component is balanced by the static friction force $f = \mu N = \mu m(g+a) \cos 	heta$.The total force exerted by the plane on the block is the vector sum (resultant) of the normal force $N$ and the friction force $f$.Therefore,the force is the resultant of $m(g+a) \cos 	heta$ normal to the plane and $\mu m(g+a) \cos 	heta$ along the plane.

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