This question involves the concepts of work done and elastic potential energy.
The work done by the spring when its extension changes from X to X/4 is "0.56 E".
The work done by the spring is equal to the elastic potential energy stored in the spring, which is given as follows:
[tex]W=E\\W=E=\frac{1}{2}kX^2---- eqn(1)[/tex]
where,
W = work done = ?
E = elastic potential energy
k = spring constant
X = extension
Now, the spring moves to an extension of X/4, so the change in extension will be:
[tex]\Delta X = X-\frac{X}{4}\\\\\Delta X = \frac{3X}{4}[/tex]
Hence, the work done will become:
[tex]W=\frac{1}{2}K\Delta X^2\\\\W=\frac{1}{2}K(\frac{3X}{4})^2\\\\W=\frac{1}{2}KX^2(0.56)\\\\using\ eqn(1):[/tex]
W = 0.56 E
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