Question

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.

1. Find the gain in the potential energy if the student is lifted straight up to the top.
2. Find the gain in the potential energy when the student climbs the stairs to the same top.
3. What do you conclude about the dependence of the potential energy on the path taken?

Answer 1

Given,

Mass (m) = 50 kg

Height (h) = 72.5 m

Acceleration due to gravity (g) = 10 m s^–2

We know,

Gain in potential energy = mgh

(i) When the student is lifted straight up to the top,

Gain in potential energy = mgh = 50 × 10 × 72.5 = 36250 J

Hence, the gain in potential energy = 36250 J.

(ii) The student gains the same height (h = 72.5 m) when they climb the stairs to the same height.

Potential energy gain = mgh = 50 × 10 × 72.5 = 36250 J

Therefore, 36250 J is the gain in potential energy.

(iii) We conclude that in both cases, the potential energy gain is the same. As a result, the potential energy is independent of the path followed and only depends on the vertical height achieved.