A) E/2
B) 2 E
C) E
D) 4 E
Correct Answer: A
Solution :
Key Idea: When a string is stretched, then work done in stretching it through a distance x is the potential energy stored in it. Potential energy stored = Work done is stretching \[or U=\frac{1}{2}k\,{{x}^{2}}\] \[Also F=k\,x\] \[or x=\frac{F}{x}\] \[So, U=\frac{1}{2}k\,{{\left( \frac{F}{k} \right)}^{2}}\] \[i.e., U\propto \,\,\frac{1}{k}(\text{for}\,\text{constant}\,\text{force})\] \[\therefore \frac{{{U}_{B}}}{{{U}_{A}}}=\frac{{{k}_{A}}}{{{k}_{B}}}\] but \[{{k}_{B}}=2{{k}_{A}}\] \[\therefore {{U}_{B}}={{U}_{A}}\times \frac{{{k}_{A}}}{2\,{{k}_{A}}}=\frac{{{U}_{A}}}{2}=\frac{E}{2}\]You need to login to perform this action.
You will be redirected in
3 sec