A block of mass m is lying on an inclined pla | vedclass

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(A) For a block of mass $m$ on an inclined plane with angle of inclination $	heta$,the forces acting on the block are:$1$. The component of weight ac

Vedclass · Accepted Answer

$mg \sin 	heta + \mu mg \cos 	heta$

Vedclass · Answer

(A) For a block of mass $m$ on an inclined plane with angle of inclination $	heta$,the forces acting on the block are:$1$. The component of weight acting down the plane: $mg \sin 	heta$.$2$. The normal force perpendicular to the plane: $N = mg \cos 	heta$.$3$. The frictional force acting down the plane (opposing the upward motion): $f = \mu N = \mu mg \cos 	heta$.To move the block up the plane,the applied force $F$ must overcome both the component of gravity and the frictional force.Therefore,$F = mg \sin 	heta + f = mg \sin 	heta + \mu mg \cos 	heta = mg(\sin 	heta + \mu \cos 	heta)$.

$A$ block of mass $m$ is lying on an inclined plane. The coefficient of friction between the plane and the block is $\mu$. The force $(F)$ required to move the block up the inclined plane will be

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