A) 400 J
B) 300 J
C) 200 J
D) 100 J
Correct Answer: B
Solution :
Potential energy stored in the spring is given \[U=\frac{1}{2}k{{x}^{2}}\] \[\therefore \] \[\frac{{{U}_{1}}}{{{U}_{2}}}={{\left( \frac{{{x}_{1}}}{{{x}_{2}}} \right)}^{2}}\] or \[\frac{100}{{{U}_{2}}}=\frac{{{(2)}^{2}}}{{{(4)}^{2}}}\] or \[{{U}_{2}}=400\,J\] \[\therefore \] Potential energy increases by \[400-100=300\text{ }J\]
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