Question

If the same fish is attached to the end of the unstretched spring and then allowed to fall from rest, through what maximum distance does it stretch the spring? (Hint: Calculate the force constant of the spring in terms of the distance d and the mass m of the fish.)

Answer 1

Answer:

x=2d

Explanation:

initial stretch in the spring is d

so using Hook's law

at equilibrium position

k×d=mg

where k= spring constant

m= mass of fish

g= acceleration due to gravity.

d=mg/k ................ (1)

in second case  by energy conservation

1/2 kx^2=mgx

x=2mg/k

using equation 1

x=2d