Respuesta :
The spring stretched 0.476 meter from equilibrium position.
The work done is given as,
[tex]workdone=\frac{1}{2}kx^{2}[/tex]
Where x is the distance stretched from equilibrium position.
Given that, [tex]k=440N/m,W=25J[/tex]
[tex]25=\frac{1}{2}*440*x^{2} \\\\x^{2} =\frac{100}{440}=0.227\\\\x=\sqrt{0.227}=0.476m[/tex]
Therefore, the spring stretched 0.476 meter from equilibrium position.
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