Final answer:
The spring constant in this scenario is approximately 2,238 N/m.
Step-by-step explanation:
To find the spring constant, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.
The formula for Hooke's Law is F = -kx, where F is the force applied, k is the spring constant, and x is the displacement.
In this case, the rock is dropped from a height of 32 m, which means it has a potential energy of mgh = 2.5 kg * 9.8 m/s^2 * 32 m = 784 J when it hits the spring.
This potential energy is converted into elastic potential energy stored in the spring, which can be calculated using the formula PE = (1/2)kx^2, where PE is the potential energy, k is the spring constant, and x is the displacement.
Since the spring is compressed 57 cm, or 0.57 m, we can substitute these values into the equation to solve for k. 784 J = (1/2)k * (0.57 m)^2.
Solving for k gives us a spring constant of approximately 2,238 N/m.