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., \[U\propto \,\,\frac{1}{k}\] (for cons \[\tan \,t\] 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