The spring extends by $x$ on loading, then energy stored by the spring is: (if $T$ is the tension in the sprin

Question

The spring extends by $x$ on loading, then energy stored by the spring is: (if $T$ is the tension in the spring and $k$ is the spring constant)

Accepted Answer

The elastic potential energy stored in a stretched spring is given by $U = \frac{1}{2}kx^2$ .
The tension $T$ in the spring is equal in magnitude to the restoring force, so $T = kx$ .
From this, the extension $x$ can be written as $x = \frac{T}{k}$.
Substituting this value of $x$ into the potential energy formula, we get:
$$U = \frac{1}{2}k \left( \frac{T}{k} \right)^2 = \frac{1}{2}k \left( \frac{T^2}{k^2} \right) = \frac{T^2}{2k}$$
Thus, the energy stored by the spring is $\frac{T^2}{2k}$.

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