Explore BrainMass

Find the total energy at the equilibrium position

Not what you're looking for? Search our solutions OR ask your own Custom question.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

A 0.50-kg mass is attached to a spring with a spring constant of 20 N/m along a horizontal, frictionless surface. The object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position. What is the total energy of the system?

https://brainmass.com/physics/energy/finding-total-energy-equilibrium-position-470027

SOLUTION This solution is FREE courtesy of BrainMass!

m = 0.5kg
k = 20 N/m
v = 1.5m/s

At the equilibrium place the spring is not stretched or compressed, hence the entire energy in the system is in the form of kinetic energy of the mass.

E = 1/2 m * v^2
=0.5*0.5*1.5^2
=0.5625 J.

Therefore, the total energy of the system is 0.5625 J.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!