Answer:
Work done in stretching a spring of force constant k through a distance \[x\], \[W=\frac{1}{2}k{{x}^{2}}\] \[\therefore \] \[\frac{{{W}_{A}}}{{{W}_{B}}}=\frac{(1/2){{k}_{A}}{{x}^{2}}}{(1/2){{k}_{B}}{{x}^{2}}}=\frac{{{k}_{A}}}{{{k}_{B}}}\] As \[{{k}_{A}}>{{k}_{B}},\] therefore, \[{{W}_{A}}>{{W}_{B}}.\]
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