You are here
-
๐ Revise, Reflect, Refine, Q5 (Page 137)
A student is slowly lifted straight up in an elevator from the ground level to the top floor of a building. Later, the same student climbs the staircase, all the way to the top. Given that the height of the building is h = 72.5 m, acceleration due to gravity is g = 10 m sโป², and student's mass is m = 50 kg.View answer
Answer-
(i) Gain in potential energy when lifted straight up
- U = mgh
- U = 50 × 10 × 72.5
- U = 36250 J
-
(ii) Gain in potential energy when climbing the stairs
- The final height is the same — 72.5 m.
- U = mgh = 50 × 10 × 72.5
- U = 36250 J
-
(iii) What do we conclude?
- The gain in potential energy is the same in both cases.
- Potential energy depends only on the height gained, NOT on the path taken.
-
(i) Gain in potential energy when lifted straight up
Q5 - A student is slowly lifted
Last updated at September 16, 2026 by Teachoo
