Q. A block of mass $5\ kg$ is on a rough horizontal surface and is at rest. Now a force of $24\ N$ is imparted to it with negligible impulse. If the coefficient of kinetic friction is 0.4 and $g = 9.8\ m/s^{2}$, then the acceleration of the block is
A) $0.26\ m/s^{2}$
B) $0.39\ m/s^{2}$
C) $0.69\ m/s^{2}$
D) $0.88\ m/s^{2}$
Explanation
(d) Net force =Applied force $-$ Friction force
$ma = 24 - \mu\ mg = 24 - 0.4 \times 5 \times 9.8 = 24 - 19.6$
$\Rightarrow a = \frac{4.4}{5} = 0.88\ m/s^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.