Final answer:
The maximum distance the spring stretches when the fish falls from rest is sqrt(2) times the equilibrium stretch distance d, derived from energy conservation principles and Hooke's Law.
Step-by-step explanation:
The student has asked about the behavior of a fish attached to a vertical spring, specifically how far the spring will stretch if the fish is allowed to fall from rest.
When the fish is slowly lowered to its equilibrium position, the system reaches a state where the force due to gravity (Fg = mg) is balanced by the spring's restoring force (Fspring = k*d), where m stands for the mass of the fish, k is the spring constant, and d is the displacement caused by the fish at equilibrium.
To calculate the spring constant, we use Hooke's Law: F = k*d, and since F equals the weight of the fish (Fg = mg), we get k = mg/d.
When the fish is dropped from the height of the unstretched spring, it converts gravitational potential energy into kinetic energy and then into elastic potential energy of the spring at the maximum stretch.
At the maximum displacement, all the gravitational potential energy has been converted into elastic potential energy. We can equate gravitational potential energy at the point of release (m*g*d) to the elastic potential energy at the maximum stretch (1/2*k*x^2), where x represents the maximum displacement.
Substituting the value of k calculated earlier gives us an equation in terms of m, g, and d that can be solved for x. The result is x = sqrt(2)*d, meaning that the spring will stretch sqrt(2) times the distance d at the maximum displacement.