Question

The world’s tallest building is the Burj Khalifa which stands at 828 m above the ground. An eccentric billionaire CEO has an office on the top floor. He insists on having a personal elevator installed that consists only of a giant spring that spans from the basement to his office when it is uncompressed. If his mass is 120 kg, what spring constant in N/m is required so that he momentarily comes to rest on the ground floor?

Answer 1

Answer:

The spring constant is approximately 2.84 N/m

Explanation:

The height of the building, h = 828 m

The mass of the billionaire that has an office on the top floor, m = 120 kg

Gravitational potential energy, P.E. = m·g·h

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

The gravitational potential energy of the billionaire at the top floor is therefore;

P.E. of billionaire at top floor = 120 kg × 9.81 m/s² × 828 m  = 974,721.6 J

The elastic potential energy of the spring, $P.E._{spring}$ is given as follows;

$P.E._{spring} = \dfrac{1}{2} \cdot k \cdot h^2$

Where;

k = The spring constant of the spring in N/m

h = The extension of the spring = The height of the building = 828 m

Given that the energy of the spring is conserved, we have;

$P.E._{spring}$ = P.E. of billionaire = 974,721.6 J

Plugging in the values gives;

$P.E._{spring} = 974,721.6 \ J = \dfrac{1}{2} \times k \times (828 \ m)^2$

Therefore;

2*974,721.6/(828^2)

$k = \dfrac{2 \times 974,721.6 \ J}{(828 \ m)^2} \approx 2.84 \ N/m$

The spring constant, k ≈ 2.84 N/m.