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(A) For a block resting on an inclined plane,the component of gravitational force acting down the plane is $F_g = mg \sin(	heta)$.Given $	heta = 30^

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(A) For a block resting on an inclined plane,the component of gravitational force acting down the plane is $F_g = mg \sin(	heta)$.Given $	heta = 30^o$,$g = 10\,m/s^2$,and the frictional force $f = 10\,N$.Since the block is at rest,the static frictional force must balance the component of gravity acting down the plane.Thus,$f = mg \sin(30^o)$.Substituting the values: $10 = m 	imes 10 	imes \sin(30^o)$.Since $\sin(30^o) = 0.5$,we have $10 = m 	imes 10 	imes 0.5$.$10 = 5m$.$m = 10 / 5 = 2\,kg$.Note: The coefficient of static friction $\mu_s = 0.8$ is provided to check if the block can remain at rest. The maximum static friction is $f_{max} = \mu_s N = \mu_s mg \cos(30^o) = 0.8 	imes 2 	imes 10 	imes 0.866 \approx 13.86\,N$. Since $10\,N < 13.86\,N$,the block remains at rest.

$A$ block rests on a rough inclined plane making an angle of $30^o$ with the horizontal. The coefficient of static friction between the block and the plane is $0.8.$ If the frictional force on the block is $10\,N,$ the mass of the block (in $kg$) is (take $g = 10\,m/s^2$)

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