Q. A block of mass 2 kg is at rest on a floor. The coefficient of static friction between block and the floor is 0.54. A horizontal force of 2.8 N is applied to the block. What should be the frictional force between the block and the floor? $(Take\ g = 10\ m/s^{2})$
A) 8.8 N
B) 5.8 N
C) 2.8 N
D) 10.8 N
Explanation
(c) $Here,\ ,\ m = 2\ kg,\ \mu_{s} = 0.54,\ F = 2.8\ N,\ g = {10\ ms}^{- 2}$
Limiting force of friction,
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$f_{s} = \mu_{s}\ R = \mu_{s}\ mg = 0.54 \times 2 \times 10 = 10.8\ N$
As $F < f_{s},$ therefore, the block dose not move. As static friction is itself an adjusting friction. Hence, frictional force between the block and the floor will be 2.8 N
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.