Q. A constant force acts on a body of mass $0.9\ kg\ $at rest for $10s.$ If the body moves a distance of $250\ m$, the magnitude of the force is
A) $3N$
B) $3.5N$
C) $4.0N$
D) $4.5N$
Explanation
(d) $u = 0,S = 250m,t = 10\sec$
$S = ut + \frac{1}{2}at^{2} \Rightarrow 250 = \frac{1}{2}a\lbrack 10\rbrack^{2} \Rightarrow a = 5m/s^{2}$
So, $F = ma = 0.9 \times 5 = 4.5N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.