Issues Determining Change in Elastic Potential Energy

[Sofa]

Homework Statement

See attached

Relevant Equations

See attached

I've attached a screengrab of the problem (Specifically, Part B, as indicated in the image) and my attempt at a solution. Summarized, my thinking was based on using $-\Delta U=\frac{Kx_i^2-Kx_f^2}{2}$.

After using up all my attempts, the solution, as it turns out, was U₂=4.91J. No variation on the above formula - or any I know, for that matter - gets me anything like that. For future reference, what exactly was I doing wrong?

 

[haruspex]

Homework Statement: See attached
Homework Equations: See attached

I've attached a screengrab of the problem (Specifically, Part B, as indicated in the image) and my attempt at a solution. Summarized, my thinking was based on using $-\Delta U=\frac{Kx_i^2-Kx_f^2}{2}$.

After using up all my attempts, the solution, as it turns out, was U₂=4.91J. No variation on the above formula - or any I know, for that matter - gets me anything like that. For future reference, what exactly was I doing wrong?View attachment 251997

You seem to have interpreted part b as being compressed (or rather, relaxed) 0.06m relative to the stretched position in part a. They mean relative to the initial unstretched position.

 

[kuruman]

The potential energy relative to the unstretched position is $U=\frac{1}{2}kx^2$ regardless of whether the spring is stretched or compressed. In part (b) you are asked for the potential energy. You calculated the change in potential energy from part (a). For future reference, read the problem carefully and asnwer what is being asked.