A 0.4 kg mass takes 8s 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 m/s²)
(Take m/s²)
Show answer and solution
Answer: 300
Solution
The object falls for 8s total. Velocity at end of 7th second: v₇ = g×7 = 70 m/s. Velocity at end of 8th second: v₈ = g×8 = 80 m/s.
Distance fallen in last second (8th second): s = v₇×1 + (1/2)g×1² = 70 + 5 = 75 m.
Loss in PE = mgh = 0.4 × 10 × 75 = 300 J.
Distance fallen in last second (8th second): s = v₇×1 + (1/2)g×1² = 70 + 5 = 75 m.
Loss in PE = mgh = 0.4 × 10 × 75 = 300 J.
[JEE Main 2023 · 29 Jan · Shift 1]