Q. 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 $g = 10\ m/s^{2}$)
A) 2.0
B) 4.0
C) 1.6
D) 2.5
Explanation
(a) Angle of repose $\alpha = \tan^{- 1}{(\mu)} = \tan^{- 1}{(0.8)} = 38.6{^\circ}$
Angle of inclined plane is given $\theta = 30{^\circ}$. It means block is at rest therefore,
Static friction = component of weight in downward direction $= mg\sin\theta = 10\ N\ \ \therefore m = \frac{10}{g \times \sin{30{^\circ}}} = 2\ kg$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.