Q. A force of $750\ N$ is applied to a block of mass $102\ kg$ to prevent it from sliding on a plane with an inclination angle $30{^\circ}$ with the horizontal. If the coefficients of static friction and kinetic friction between the block and the plane are 0.4 and 0.3 respectively, then the frictional force acting on the block is
A) $750\ N$
B) $500\ N$
C) $345\ N$
D) $250\ N$
Explanation
(d) Net force along the plane
$= P - mg\sin\theta = 750 - 500 = 250\ N$
Limiting friction $= F_{1} = \mu_{s}R = \mu_{s}mg\cos\theta$
$= 0.4 \times 102 \times 9.8 \times \cos 30 = 346\ N$
As net external force is less than limiting friction therefore friction on the body will be $250\ N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.