Calculating Total Energy of Block-Spring System

[trickymax301]

Question: A block of mass .05 kg is pulled .3 m from its equilibrium position and released. The spring constant is 5 N/m

What is the total energy of the block-spring system?

My book says E = K + U or E = (1/2)mv^2 + (1/2)kx^2. My book also says E_total = (1/2)kA(amplitude)^2. Which formula is correct? I've just solved for the E_total equation and I found the answer to be .75 J. Can anyone confirm this answer for me? My work is below...

(1/2)(5)(.3^2) = .75 J

Thanks

 

[Integral]

How did you get .75J from that? Try again.

 

[trickymax301]

i lied. .225 J

Thanks for checking for me. I forgot to square the .3

 

[Integral]

With that cleared up.

The 2 relationships you have posted.

$$E = K + U = \frac 1 2 m v^2 + \frac 1 2 k x^2$$

and

$$E = \frac 1 2 k A^2$$

are the same. The amplitude is defined as when the displacement is at its maximum so A = x. At the point in time when the displacement is at its maximum, the velocity is zero. So in the first relationship, let x=A and v=0.