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 $(Take\ g = 10\ {ms}^{- 2})$
A) 2.0
B) 4.0
C) 1.6
D) 2.5
Explanation
(a) Let the mass of block be $m$
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Friction force is rest position
$F = mg\sin{30{^\circ}}$
(This is static frictional force and may be less than the limiting frictional force)
$10 = m \times 10 \times \frac{1}{2}$
$\therefore\ \ m = \frac{2 \times 10}{10} = 2\ kg$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.