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Two identical springs have the same force constant of 147 Nm-1 . What elongation will be produced in each case shown in figure?

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Here k = 147 Nm-1

In figure (a), the effective spring constant,

k1 = k + k = 2k = 2 × 147 = 294 Nm-1

∴ Elongation in the spring,

y1 = \(\frac{mg}{k}=\frac{5\times9.8}{294}=\frac{1}{6}m\) 

In figure (b), the effective spring constant,

\(k_2=\frac{k\times k}{k\times k}=\frac{k}{2}\) 

= \(\frac{147}{2}\)m-1 

∴ Total elongation in the spring

\(y_2=\frac{5\times 9.8\times 2}{147}=\frac{2}{3}\) m

In figure (c), the effective constant,

k= 147 Nm-1

∴ Elongation in the spring,

y3 = \(\frac{5\times 9.8}{147}=\frac{1}{3}\) m.

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