Q 11-05-089JEE MainJEE Main 2023 (29 Jan, Shift 1)Easy
A $0.4\ \text{kg}$ mass takes $8\ \text{s}$ to reach ground when dropped from a certain height $P$ above surface of earth. The loss of potential energy in the last second of fall is ______ J. [Take $g=10\ \text{m s}^{-2}$]
Numerical value type. Enter your answer.
Answer: 300
Distance fallen in $t$ seconds: $s=\tfrac12gt^2$.
Distance in the last (8th) second: $s_8-s_7=\tfrac12\times10\,(64-49)=75\ \text{m}$.
Loss of PE $=mgh=0.4\times10\times75=300\ \text{J}$.
Solution by Sreeraj P, M.Sc Physics