A block of mass m is on an inclined plane of | vedclass

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(A) For a block on an inclined plane,the force of gravity component down the plane is $mg \sin 	heta$. The maximum static frictional force is $f_{max

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(A) For a block on an inclined plane,the force of gravity component down the plane is $mg \sin 	heta$. The maximum static frictional force is $f_{max} = \mu mg \cos 	heta$. When a force $P$ is applied parallel to the plane,the equation of equilibrium is $P + f - mg \sin 	heta = 0$,where $f$ is the static friction force. Thus,$f = mg \sin 	heta - P$. This is a linear equation of the form $y = -mx + c$,which represents a straight line with a negative slope. At $P = P_1 = mg(\sin 	heta - \mu \cos 	heta)$,$f = mg \sin 	heta - mg(\sin 	heta - \mu \cos 	heta) = \mu mg \cos 	heta$. At $P = P_2 = mg(\sin 	heta + \mu \cos 	heta)$,$f = mg \sin 	heta - mg(\sin 	heta + \mu \cos 	heta) = -\mu mg \cos 	heta$. As $P$ increases from $P_1$ to $P_2$,the frictional force $f$ decreases linearly from $\mu mg \cos 	heta$ to $-\mu mg \cos 	heta$. This corresponds to the graph shown in option $A$.

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