A student is slowly lifted straight up in an elevator from the ground level. Class 9

A student is slowly lifted straight up in an elevator from the ground level. Class 9

Question 1.

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 ms^-2, and student’s mass is m = 50 kg. Class 9

(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?

Answer:

(i) PE gain in elevator: U = mgh = 50 × 10 × 72.5 = 36250 J.

(ii) PE gain climbing stairs: U = mgh = 50 × 10 × 72.5 = 36250 J.

(iii) Conclusion: The gain in potential energy is the same in both cases (36250 J), even though the paths are different. This shows that gravitational potential energy depends only on the height gained (h), not on the path taken. Potential energy is a path-independent quantity.