A 3.5 kg object is suspended from a spring of spring constant 725 N/m. Starting from the equilibrium position, the object has an initial speed of 7.9 m/s. What is the maximum displacement of the object during its subsequent motion? (In cm)

Answer 1

Kinetic energy at start = (1/2) m v^2
= (1/2)(3.5)(7.9)^2
= 109 Joules

At the maximum displacement, it stops, so that kinetic energy has all gone into stretching the spring which stores it as potential energy
Potential energy in spring = (1/2) k x^2
so
(1/2) ( 725) * x^2 = 109 Joules
x^2 = .301
x = .549 meters
= 55 cm