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(A) Let the mass of the block be $m \, kg$.The block is at rest on the inclined plane.The component of the gravitational force acting down the plane i

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(A) Let the mass of the block be $m \, kg$.The block is at rest on the inclined plane.The component of the gravitational force acting down the plane is $mg \sin 30^o$.Since the block is in equilibrium,the static frictional force $F$ must balance this component:$F = mg \sin 30^o$Given $F = 10 \, N$ and $g = 10 \, m/s^2$,we have:$10 = m 	imes 10 	imes \sin 30^o$$10 = 10m 	imes \frac{1}{2}$$10 = 5m$$m = \frac{10}{5} = 2 \, kg$.(Note: The maximum static friction is $f_{max} = \mu N = \mu mg \cos 30^o = 0.8 	imes 2 	imes 10 	imes \frac{\sqrt{3}}{2} \approx 13.86 \, N$. Since $10 \, N < 13.86 \, N$,the block indeed 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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