Calculating Total Energy of Block-Spring System

AI Thread Summary
The total energy of a block-spring system can be calculated using the formula E = (1/2)kA^2, where A is the amplitude of the displacement. In this case, a block of mass 0.05 kg is pulled 0.3 m from equilibrium, and the spring constant is 5 N/m. The initial calculation of total energy was incorrectly stated as 0.75 J due to a miscalculation, which was later corrected to 0.225 J. The two energy relationships provided, E = K + U and E = (1/2)kA^2, are equivalent when considering maximum displacement. Understanding that at maximum displacement, the velocity is zero clarifies the relationship between the two formulas.
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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
 
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How did you get .75J from that? Try again.
 
i lied. .225 J

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