A) \[8.7\times {{10}^{-2}}J\]
B) \[12.2\times {{10}^{-2}}J\]
C) \[8.7\times {{10}^{-1}}J\]
D) \[12.2\times {{10}^{-1}}J\]
Correct Answer: A
Solution :
Key Idea Work done is equal to difference in potential energies for two different positions of spring. Given, \[F=-5x-16\,{{x}^{3}}\] \[=-(5+16{{x}^{2}})x\] \[=-kx\] where \[k(=5+16{{x}^{2}})\] is force constant of spring. Therefore, work done in stretching the spring from position \[{{x}_{1}}\] to position \[{{x}_{2}}\] is \[W=\frac{1}{2}{{k}_{2}}x_{2}^{2}-\frac{1}{2}{{k}_{1}}x_{1}^{2}\] we have, \[{{x}_{1}}=0.1\,m\] and \[{{x}_{2}}=0.2\,m\]. \[\therefore \] \[W=\frac{1}{2}\,[5+16\,{{(0.2)}^{2}}]\,{{(0.2)}^{2}}\] \[-\frac{1}{2}[5+16\,{{(0.1)}^{2}}]\,\,{{(0.1)}^{2}}\] \[=2.82\times 4\times {{10}^{-2}}-2.58\times {{10}^{-2}}\] \[=8.7\times {{10}^{-2}}J\]
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