## KE of a Spring

A mass-on-a-spring system has a spring constant k = 37.5 N/m and a mass m = 0.35 kg. If it's given an initial displacement of 0.415 m from the equilibrium position and then released, what is the maximum kinetic energy of the oscillator?

**3.23 J**## **Approach**

This is a conservation of energy problem. At the maximum displacement of 0.415 m the mass is at rest and all the energy is in the potential energy of the spring. The maximum kinetic energy occurs at the equilibrium position. At this position the potential energy is zero and all the energy is kinetic.

**Solution**

Given quantities:

KE=PE+1/2kA^2

Putting in the numbers...

KE=1/2(37.5N/m)(0.415m)^2=3.23 J

- mass, m = 0.35 kg
- spring constant, k = 37.5 N/m
- amplitude A = 0.415 m

KE=PE+1/2kA^2

Putting in the numbers...

KE=1/2(37.5N/m)(0.415m)^2=3.23 J