A) \[\frac{1}{2}k{{l}_{1}}\left( 2l+{{l}_{1}} \right)\]
B) \[\frac{1}{2}k{{l}_{1}}^{2}\]
C) \[\frac{1}{2}k\left( {{l}^{2}}+{{l}_{1}}^{2} \right)\]
D) \[\frac{1}{2}k\left( {{l}^{2}}_{1}-{{l}_{1}}^{2} \right)\]
Correct Answer: D
Solution :
\[\operatorname{Work} done in stretching = {{U}_{2}}- {{U}_{1}}\] \[\operatorname{U}\left( potential energy \right) = \frac{1}{2}k{{x}^{2}}\] \[\therefore x =\]extension from natural length \[{{\operatorname{U}}_{2}}=\frac{1}{2}\times K{{\ell }^{2}}_{1}{{\operatorname{U}}_{1}}=\frac{1}{2}\times K{{\ell }^{2}}\] \[\operatorname{Work} done =\frac{k}{2}[{{\ell }^{2}}_{1}-{{\ell }^{2}}]\]
You need to login to perform this action.
You will be redirected in
3 sec
