A) \[8.7\times {{10}^{-2}}J\]
B) \[12.2\times {{10}^{-2}}J\]
C) \[8.1\times {{10}^{-1}}J\]
D) \[12.2\times {{10}^{-1}}J\]
Correct Answer: A
Solution :
Spring force is given by \[F=(-5x-16{{x}^{3}})N\] or \[F=(-5+16{{x}^{2}})x\] ...(i) but we are familiar with linear restoring force, \[F=-kx\] ...(ii) where k is force constant of spring. Comparing Eqs. (i) and (ii), we have \[k=5+16{{x}^{2}}\] Therefore, work done in stretching the spring from position\[{{x}_{1}}\]to\[{{x}_{2}}\]is, \[W=\frac{1}{2}{{k}_{2}}x_{2}^{2}-\frac{1}{2}{{k}_{1}}x_{1}^{2}\] ?..(iv) Given, \[{{x}_{1}}\ =0.1m\,and\,x=0.2m\] So, \[W=\frac{1}{2}[5+16{{(0.2)}^{2}}]{{(0.2)}^{2}}\] \[-\frac{1}{2}[5+16{{(0.1)}^{2}}]{{(0.1)}^{2}}\] \[=2.82\times 4\times {{10}^{-2}}-2.58\times {{10}^{-2}}\] \[=8.7\times {{10}^{-2}}J\]
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