Q. A block of mass $5\ kg$ is moving horizontally at a speed of $1.5\ m/s$. A perpendicular force of $5N$ acts on it for $4\ sec$. What will be the distance of the block from the point where the force started acting
A) $10\ m$
B) $8\ m$
C) $6\ m$
D) $2\ m$
Explanation
(a) $S_{Horizontal} = ut = 1.5 \times 4 = 6m$
$S_{Vertical} = \frac{1}{2}at^{2} = \frac{1}{2}\frac{F}{m}t^{2} = \frac{1}{2} \times 1 \times 16 = 8\ m$
$S_{Net} = \sqrt{6^{2} + 8^{2}} = 10\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.