Exercise 7.5
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 = m, acceleration due to gravity is g = m , and student’s mass is m = 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?
NCERT’s answer
(i)
$\displaystyle 36250$ J (ii) $\displaystyle 36250$ J (iii) does not depend upon the path
(i) Gain in potential energy in the lift = $\displaystyle 36250$ J.
\(\displaystyle U = mgh = 50\ \text{kg} \times 10\ \text{m s}^{-2} \times 72.5\ \text{m} = 36250\ \text{J} \)
(ii) Gain in potential energy up the stairs = $\displaystyle 36250$ J — exactly the same.
\(\displaystyle U = mgh = 50\ \text{kg} \times 10\ \text{m s}^{-2} \times 72.5\ \text{m} = 36250\ \text{J} \)
The staircase is longer than the lift shaft, but the height gained is the same $\displaystyle 72.5$ m, and only the height enters \(\displaystyle mgh \).
(iii) Gravitational potential energy does not depend on the path taken — only on the vertical height between the starting and finishing points.