A) \[2\times {{10}^{-2}}J\]
B) \[4\times {{10}^{-2}}J\]
C) \[8\times {{10}^{-2}}J\]
D) \[16\times {{10}^{-2}}J\]
Correct Answer: A
Solution :
Maximum KE of the system = Maximum PE of the system \[=\frac{1}{2}k{{x}^{2}}\] \[=\frac{1}{2}\times 16\times {{(5\times {{10}^{-2}})}^{2}}\] \[=2\times {{10}^{-2}}J\]
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