Answer:
(a) As \[\text{W}=\frac{1}{2}\text{K}{{\text{x}}^{\text{2}}}\], and \[{{\text{K}}_{\text{A}}}>{{\text{K}}_{\text{B}}}\]
work done on spring A will be more than work done on spring B (for
same x)
As F = Kx, therefore, x = F/K
As \[{{\text{K}}_{\text{A}}}>{{\text{K}}_{\text{B}}}\]
\[{{\text{X}}_{\text{A}}}<{{\text{X}}_{\text{B}}}\] As \[\text{W}=\frac{1}{2}\text{K}{{\text{x}}^{\text{2}}}=\frac{1}{2}\left(
\frac{F}{x} \right){{x}^{2}}=\frac{1}{2}Fx\]
As \[{{\text{x}}_{\text{A}}}<{{\text{x}}_{\text{B}}}\] \[{{\text{W}}_{\text{A}}}<{{\text{W}}_{\text{B}}}\] or \[{{\text{W}}_{\text{B}}}>{{\text{W}}_{\text{A}}}\]
i.e. more work is required in case of spring B than in the case of
spring A.
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