Question

A 2.0 kg box moves back and forth on a horizontal frictionless surface between two different springs with one 32 N/cm and the other at 16 N/cm.The box is initially pressed against the stronger spring, compressing it 3.5 cm , and then is released from rest.

By how much will the box compress the weaker spring? What is the maximum speed the box will reach?

Answer 1

potential energy U = (1/2) k x^2
(1/2)(3200 N/m)(.035)^2

max speed = Vmax
(1/2) m Vmax^2 = U

compression of weak spring, same potential energy when stopped
U = (1/2)(1600)(x^2)
x is in meters of course