A) \[\frac{1}{2}Fx\]
B) \[\frac{1}{2}Yx\]
C) \[\frac{1}{2}F{{x}^{2}}\]
D) none of these
Correct Answer: A
Solution :
When a wire is stretched through a length, then work has to be done, this work is stored in the wire in the form of elastic potential energy. Potential energy of stretched wire is \[U=\frac{1}{2}\times stress\times strain\] \[\therefore \] \[U=\frac{1}{2}\times F\times x\] \[\Rightarrow \] \[U=\frac{1}{2}Fx\]You need to login to perform this action.
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