Q.
A block of weight 5N is pushed against a vertical wall by a force 12N. The coefficient of friction between the wall and block is 0.6. The magnitude of the force exerted by the wall on the block is
A) 12 N
B) 5 N
C) 7.2 N
D) 13 N
Explanation
(d) Wall applies 2 forces of the block (i) normal reaction, $R = 12$ N, and (ii) frictional force, $f_{2} = mg = 5$ N tangentially upward
[image:5_image266.emf]
$\therefore$ Total force exerted by wall on block
$F = \sqrt{N^{2} + f_{s}^{2}} = \sqrt{(12)^{2} + (5)^{2}} = 13$N
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.