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 g = 10 m/s²)

Solution

Physics sir · 4 steps

Do not touch the full height: find only the distance covered in the 8th second, then use mgh.

1. Distance in 8th second. The body starts from rest, so the distance covered in the nth second is g/2(2n-1). Put n = 8.

   s₈ = 10/2(2 × 8 - 1) = 5 × 15 = 75 m

2. Check another way. In 8 s it falls 1/2 × 10 × 8² = 320 m. In 7 s it falls 1/2 × 10 × 7² = 245 m. The difference is 320 - 245 = 75 m, the same answer.
3. Loss of PE. The potential energy lost over this 75 m drop is mgh = 0.4 × 10 × 75 = 300 J.
4. So. The loss of potential energy in the last second is 300 J.

The common slip: 1280. Using the full height of 320 m in mgh gives 1280 J, which is the loss over the whole 8 s, not over the last second.

The idea to keep. For anything that happens in a particular second of free fall, first find the distance in that second, g/2(2n-1) from rest, then use it in mgh.

Answer from the NTA final answer key. Solution by PrepWiser.