Final answer:
To design a safety device for a temporary lift (elevator) that meets certain conditions, we need to calculate the spring constant needed for the spring. By setting the potential energy at the starting position equal to the potential energy at the desired stopping position and solving for the spring constant, we can determine the required value. The spring constant needed for the spring is approximately 3036.18 N/m.
Step-by-step explanation:
To design a safety device for the temporary lift (elevator), we need to determine the spring constant needed for the spring. Given the elevator car's mass (m = 571 kg), starting height (yᵢ = 3.27 m above the top of the uncompressed spring), initial velocity (vᵢ = 6.06 m/s), and the desired stopping distance (y = 2.02 m below the original position of the spring), we can calculate the needed spring constant (k).
The potential energy of the elevator car at its starting position is given by m * g * yᵢ (where g is the acceleration due to gravity, approximately 9.8 m/s²). The potential energy at the desired stopping position is m * g * y. The difference in potential energy is equal to the work done by the spring, which is given by 0.5 * k * Δy² (where Δy = y - yᵢ). Setting these two energies equal to each other, we can solve for k.
Using the equation:
m * g * yᵢ = 0.5 * k * Δy² + m * g * y
Substituting the given values, we can solve for k:
571 kg * 9.8 m/s² * 3.27 m = 0.5 * k * (2.02 m)² + 571 kg * 9.8 m/s² * 2.02 m
Simplifying the equation, we find:
k ≈ 3036.18 N/m