A) 20J
B) 25J
C) 5J
D) 15J
Correct Answer: B
Solution :
Initial potential energy\[=\frac{1}{2}k{{x}^{2}}=20J\] Final potential energy\[=\frac{1}{2}k{{\left( x+\frac{x}{2} \right)}^{2}}\] \[=\frac{9}{4}.\frac{1}{2}k{{x}^{2}}=\frac{9}{4}\times 20=45J\] So, work done = change in potential energy of spring \[=45-20=25J\]You need to login to perform this action.
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