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 :
Force exerted on spring is given by \[F=(-5x-16{{x}^{3}})N\] or \[F=(5+16{{x}^{2}})x\,N\] ... (i) Comparing Eq. (i) with \[F=(5+16{{x}^{2}})x\,N\] where k is a force constant. We have \[k=5+16{{x}^{2}}\] Work done in stretching the spring from \[{{x}_{1}}=0.1\,m\] to \[{{x}_{2}}=0.2\,m\] is \[W=\frac{1}{2}\,{{k}_{2}}x_{2}^{2}-\frac{1}{2}{{k}_{1}}x_{1}^{2}\] \[W=\frac{1}{2}[5+16x_{2}^{2}]x_{2}^{2}-\frac{1}{2}[5+16x_{1}^{2}]\,x_{1}^{2}\] Substituting the given values, we obtain \[W=\frac{1}{2}[5+16{{(0.2)}^{2}}]\,{{(0.2)}^{2}}\] \[-\frac{1}{2}[5+16\,{{(0.1)}^{2}}]\,{{(0.1)}^{2}}\] \[=2.82\times 4\times {{10}^{-2}}-2.58\times 1\times {{10}^{-2}}\] \[=8.7\times {{10}^{-2}}J\]You need to login to perform this action.
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