Answer:
(a) For a given spring, \[\text{F}=\text{kx}\]
\[\therefore \] \[\frac{{{F}_{2}}}{{{F}_{1}}}=\frac{k{{x}_{2}}}{k{{x}_{1}}}=\frac{2x}{x}=2\]
(b) For a given spring, \[\text{U}=\frac{1}{2}\text{k}{{\text{x}}^{\text{2}}}\] \ \[\frac{{{U}_{2}}}{{{U}_{1}}}=\frac{\frac{1}{2}kx_{2}^{2}}{\frac{1}{2}kx_{1}^{2}}=\frac{{{(2x)}^{2}}}{{{x}^{2}}}=4\]
(c) As work done in stretching the spring is stored in the spring
in the form of elastic potential energy of the spring, therefore, \[\text{W}={{\text{U}}_{\text{2}}}-{{\text{U}}_{\text{1}}}=\frac{1}{2}kx_{2}^{2}-\frac{1}{2}kx_{1}^{2}=\frac{1}{2}k[{{(2x)}^{2}}-{{x}^{2}}]=\frac{3}{2}k{{x}^{2}}\]
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