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?
To find the spring constant required for the eccentric billionaire CEO to momentarily come to rest on the ground floor, we can use Hooke's Law:
F = -kx
Where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.
In this scenario, the weight of the CEO needs to be balanced by the force exerted by the spring when he momentarily comes to rest on the ground floor.
Weight (W) = mass (m) × acceleration due to gravity (g)
W = 120 kg × 9.8 m/s²
W = 1176 N
At equilibrium, the force exerted by the spring must be equal to the weight of the CEO:
-kx = -1176 N
Since the CEO is momentarily at rest, his displacement from the equilibrium position is the distance the spring is compressed, which is the height of the Burj Khalifa (828 m).
k × 828 m = 1176 N
Simplifying the equation, we have:
k = 1176 N / 828 m
k ≈ 1.42 N/m
Therefore, a spring constant of approximately 1.42 N/m is required for the billionaire CEO to momentarily come to rest on the ground floor of the Burj Khalifa.