Answer:
[tex]L'=12.0918\ m[/tex]
Explanation:
Given:
length of unstrained spring, [tex]l=12\ cm=0.12\ m[/tex]
mass attached to the spring, [tex]m=3.15\ kg[/tex]
length of the spring after hanging the mass, [tex]L=13.3\ cm=0.133\ m[/tex]
energy to be stored in the spring, [tex]E=10\ J[/tex]
The length of the spring stretched after hanging the mass:
[tex]\delta l=L-l[/tex]
[tex]\delta l=0.133-0.12[/tex]
[tex]\delta l=0.013\ m[/tex]
Therefore the stiffness constant of the spring:
[tex]k=\frac{F}{\delta l}[/tex]
where
F = is the force to which the spring is subjected for the corresponding change in length [tex]\delta l[/tex].
here, [tex]F=m.g[/tex]
[tex]k=\frac{3.15\times 9.8}{0.013}[/tex]
[tex]k=2374.6\ N.m^{-1}[/tex]
As we know that the spring potential energy is given as:
[tex]E=\frac{1}{2} \times k.\delta L^2[/tex]
where
[tex]\delta L =[/tex] extension in the length of the spring
[tex]10=0.5\times 2374.6\times (\delta L)^2[/tex]
[tex]\delta L=0.0918\ m[/tex]
So the total length of the spring when it stores 10 J energy:
[tex]L'=l+\delta L[/tex]
[tex]L'=12+0.0918[/tex]
[tex]L'=12.0918\ m[/tex]
The total length of the spring when the given energy is stored is 12.84 cm.
The given parameters;
The extension of the spring is calculated as follow;
Δl = l₂ - l₁
Δl = 13.3 cm - 12 cm
Δl = 1.3 cm
The spring constant is calculated as follows;
[tex]F = kx\\\\mg = kx\\\\k = \frac{mg}{x} \\\\k = \frac{3.15 \times 9.8}{0.013} \\\\k = 2,374.62 \ Nm[/tex]
The extension on the spring when 10 J is stored in the spring is calculated as follows;
[tex]U = \frac{1}{2} kx^2\\\\2U = kx^2\\\\x^2 = \frac{2U}{k} \\\\x = \sqrt{\frac{2U}{k}} \\\\x = \sqrt{\frac{2\times 10}{2374.62}} \\\\x = 0.0084 \ m\\\\x = 0.84 \ cm[/tex]
The total length of the spring = 0.84 cm + 12 cm
= 12.84 cm.
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