A) 12.50 N-m
B) 18.75 N-m
C) 25.00 N-m
D) 6.25 N-m
Correct Answer: B
Solution :
\[{{W}_{1}}=\frac{1}{2}k\times x_{1}^{2}\] \[=\frac{1}{2}\times 5\times {{10}^{3}}\times {{(5\times {{10}^{-2}})}^{2}}\] \[=6.25\,\,J\] \[{{W}_{2}}=\frac{1}{2}k{{({{x}_{1}}+{{x}_{2}})}^{2}}\] \[=\frac{1}{2}\times 5\times {{10}^{3}}{{(5\times {{10}^{2}}+5\times {{10}^{-2}})}^{2}}\] \[=25\,\,J\] Net work done\[={{W}_{2}}-{{W}_{1}}\] \[=25-6.25\] \[=18.75\,\,J=18.75\,\,N-m\]
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