A block of mass m is released on a smooth inc | vedclass

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(D) The force exerted by the plane on the block is the normal reaction force $N$.When a block of mass $m$ is placed on an inclined plane with an angle

Vedclass · Accepted Answer

$m g \cos 	heta$

Vedclass · Answer

(D) The force exerted by the plane on the block is the normal reaction force $N$.When a block of mass $m$ is placed on an inclined plane with an angle of inclination $	heta$,the gravitational force $mg$ acts vertically downwards.We can resolve the gravitational force into two rectangular components:$1$. $A$ component perpendicular to the inclined plane: $mg \cos 	heta$.$2$. $A$ component parallel to the inclined plane: $mg \sin 	heta$.Since the block does not move in the direction perpendicular to the plane,the normal force $N$ exerted by the plane must balance the perpendicular component of the gravitational force.Therefore,$N = mg \cos 	heta$.Thus,the magnitude of the force exerted by the plane on the block is $mg \cos 	heta$.

$A$ block of mass $m$ is released on a smooth inclined plane of inclination $\theta$ with the horizontal. The force exerted by the plane on the block has a magnitude

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