Q. A block of mass $M = 5\ kg$ is resting on a rough horizontal surface for which the coefficient of friction is 0.2. When a force $F = 40\ N$ is applied, the acceleration of the block will be $(g = 10\ ms^{2})$
A) $5.73\ m/\sec^{2}$
B) $8.0\ m/\sec^{2}$
C) $3.17\ m/\sec^{2}$
D) $10.0\ m/\sec^{2}$
Explanation
(a) Kinetic friction $= \mu_{k}R = 0.2(mg - F\sin{30{^\circ}})$
$= 0.2\left( 5 \times 10 - 40 \times \frac{1}{2} \right) = 0.2(50 - 20) = 6N$
Acceleration of the block = $\frac{F\cos{30{^\circ}} - Kinetic\ friction}{mass}$
$= \frac{40 \times \frac{\sqrt{3}}{2} - 6}{5} = 5.73\ m/s^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.