Q. A block of mass 2 kg is placed on the floor. The coefficient of static friction is 0.4. If a force of 2.8 N is applied on the block parallel to floor, the force of friction between the block and floor (Taking $g = 10\ ms^{- 2}$) is
A) 2.8 N
B) 8 N
C) 2 N
D) Zero
Explanation
(a) Minimum force required to move the block
$= \mu R = \mu mg = 0.4 \times 2 \times 10 = 8$ N
Since the force applied is only 2.8 N, the block fails to move and static fraction = applied force = 2.8 N
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.