# Energy in Simple Harmonic Motion

1. Aug 23, 2009

Problem:

A 2.4-kg object on a frictionless horizontal surface is attached to one end of a horizontal spring of force constant k=4.5 kN/m. The other end of the spring is held stationary. The spring is strecthed 10 cm from equilibrium and released. find the system's total mechanical energy.

Attempt at solution:

It seems as if I should be able to just substitute values into the equation $$E_{sys}=\frac{1}{2}kA_o^2$$ where $$A_o$$ is just the amplitude of the motion. The given units seem inconsistent, I need to convert k into centimeters by mulitplying by the factor $$\frac{1m}{100cm}$$ which gives $$k=0.045 kN/cm$$. I get 2.25 joules. My book gives an answer of 23 joules. Where am I going wrong?

2. Aug 23, 2009

### Tom Mattson

Staff Emeritus
If you convert the spring constant to $kN/cm$ then your answer will be in $kN\cdot cm$, which is not the same as Joules. If you convert your answer to Joules you'll have it right.

But it would be a lot easier to convert $A_0$ to meters instead...