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 $h = 72.5 \mathrm{~m}$, acceleration due to gravity is $g = 10 \mathrm{~ms^{-2}}$, and the student's mass is $m = 50 \mathrm{~kg}$.

(A) Find the gain in the potential energy if the student is lifted straight up to the top.

(B) Find the gain in the potential energy when the student climbs the stairs to the same top.

(C) What do you conclude about the dependence of the potential energy on the path taken?

Given:\n$h = 72.5 \,\mathrm{m}$\n$g = 10 \,\mathrm{m} \cdot \mathrm{s}^{-2}$\n$m = 50 \,\mathrm{kg}$\n(A) Gain in potential energy when lifted by elevator:

$$\text{Potential energy} = mgh$$

$$= 50 \times 10 \times 72.5$$

$$= 36250 \text{ J}$$

Therefore, the gain in potential energy is 36250 J.

(B) Gain in potential energy when climbing the stairs:

$$\text{Potential Energy} = mgh$$

$$= 50 \times 10 \times 72.5$$

$$= 36250 \text{ J}$$

Therefore, the gain in potential energy is 36250 J.

(C) The gain in potential energy is the same in both cases. Therefore, it is concluded that the potential energy depends only on the initial and final heights and not on the path taken.