Q 11-04-103JEE MainJEE Main 2023 (29 Jan, Shift 2)Easy
Force acts for $20\ \text{s}$ on a body of mass $20\ \text{kg}$, starting from rest, after which the force ceases and then body describes $50\ \text{m}$ in the next $10\ \text{s}$. The value of force will be
Answer: (B) $5\ \text{N}$
After the force stops the body moves uniformly: $v=\dfrac{50}{10}=5\ \text{m s}^{-1}$.
This speed was gained in $20\ \text{s}$ from rest: $a=\dfrac{5}{20}=0.25\ \text{m s}^{-2}$, so $F=ma=20\times0.25=5\ \text{N}$.
Solution by Sreeraj P, M.Sc Physics