Aproud deep-sea fisherman hangs a 65.0-kg fish from an ideal spring having negligible mass. the fish stretches the spring 0.180 m. a)find the force constant of the spring. express your answer with the appropriate units. b)the fish is now pulled down 5.00 cm and released. what is the period of oscillation of the fish? express your answer with the appropriate units. c)what is the maximum speed it will reach? express your answer with the appropriate units.

(a) 3539 N/m

Hook's law states that:

where F is the force applied on the spring, k is the spring constant, x is the stretching of the spring.

In this problem, we have:

is the force applied (the weight of the fish)

is the stretching of the spring

Solving the equation for k, we find the spring constant:

(b) 0.85 s

The period of oscillation of a spring-mass system is

In this case,

m = 65.0 kg

k = 3539 N/m

Substituting into the formula,

(c) 0.37 m/s

The initial elastic potential energy of the spring when the fish is pulled down is:

where

is the stretching of the spring with respect to the initial position

Substituting,

The spring reaches its maximum speed when it crosses the equilibrium position, for which , so when all the elastic potential energy has been converted into kinetic energy:

where v is the speed of the fish. Solving for v, we find

i believe the answer is a

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge