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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 s⁻², and student's mass is m = 50 kg. (i) Find the gain in the potential energy if the student is lifted straight up to the top. (ii) Find the gain in the potential energy when the student climbs the stairs to the same top. (iii) What do you conclude about the dependence of the potential energy on the path taken?
Show Answer
(i) \(U = mgh = 50 \times 10 \times 72.5 = 36250\ \text{J}\).
(ii) Climbing the stairs reaches the same height, so the gain is the same : \(36250\ \text{J}\).
(iii) The gain in potential energy depends only on the vertical height , not on the path taken.
