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(A) For a block at rest on an inclined plane,the component of gravitational force acting down the plane is balanced by the static frictional force act

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) For a block at rest on an inclined plane,the component of gravitational force acting down the plane is balanced by the static frictional force acting up the plane.$f_s = mg \sin 	heta$Given:$f_s = 10 \, N$$	heta = 30^{\circ}$$g = 10 \, m/s^2$Substituting the values:$10 = m 	imes 10 	imes \sin 30^{\circ}$$10 = m 	imes 10 	imes 0.5$$10 = 5m$$m = \frac{10}{5} = 2 \, kg$Thus,the mass of the block is $2 \, kg$.

$A$ block rests on a rough inclined plane making an angle of $30^{\circ}$ 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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