If a spring extends by `x` on loading, then the energy stored by the spring is (if T is tension in the spring and k is spring constant)

Options

A. `(T^(2))/(2k)`
B. `(T^(2))/(2k^(2))`
C. `(2k)/(T^(2))`
D. `(2T^(2))/(k)`

Answer

A

Explanation

`U=(F^(2))/(2k)=(T^(2))/(2k)`