A) \[12.50\]N-m
B) \[18.75\]N-m
C) \[25.00\] N-m
D) \[6.25\] N-m
Correct Answer: B
Solution :
When a spring is deformed (whether it is stretched or compressed) the elastic potential energy is stored in the spring. The magnitude of this energy is \[\frac{1}{2}k{{x}^{2}}\]. If the deformation is more amount of elastic potential energy stored in the spring will b6 more. Work required to stretch the spring by 5cm, \[{{W}_{1}}=\frac{1}{2}k\times {{x}_{1}}^{2}=\frac{1}{2}\times 5\times {{10}^{3}}{{(5\times {{10}^{-2}})}^{2}}\] = 6.25 J Work required to further stretch the spring by 5cm, \[{{W}_{2}}=\frac{1}{2}k\,{{({{x}_{1}}+{{x}_{2}})}^{2}}\] \[=\frac{1}{2}\times 5\times {{10}^{3}}{{(5\times {{10}^{-2}}+5\times {{10}^{-2}})}^{2}}\] = 25 J Net work done \[={{W}_{2}}- & {{W}_{1}}\] = 25 - 6.25 = 18.75 J = 18.75 N-m
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