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.
Find the gain in the potential energy if the student is lifted straight up to the top.
36250 J
Find the gain in the potential energy when the student climbs the stairs to the same top.
36250 J
What do you conclude about the dependence of the potential energy on the path taken?
Potential energy does not depend on the path taken; it depends only on the vertical height from the reference level.
Explanation
The potential energy of an object at a height is given by PE = mgh, where m is mass, g is acceleration due to gravity, and h is the vertical height. Since both scenarios involve the same student reaching the same vertical height, the potential energy gain is identical in both cases. This demonstrates that gravitational potential energy is a function of vertical displacement only, not the path traversed.