A spring with a spring constant of 7 N/m is stretched until extended by 0.7 m; how much elastic potential energy is stored in the spring?

What you are being asked to find

When a spring is stretched, work is done against the spring force. That work is stored as elastic potential energy, which depends on the spring constant and the extension.

Using the spring energy formula

Elastic potential energy: $$E=\tfrac12 kx^2$$ Here, $k=7\,\text{N/m}$ and the extension is $x=0.7\,\text{m}$.

Substituting values and calculating

$$E=\tfrac12(7)(0.7)^2$$ Compute $x^2$ first: $$0.7^2=0.49$$ Now multiply: $$E=0.5\times 7\times 0.49=1.715\,\text{J}$$ Rounded to 2 decimal places: $$E\approx 1.72\,\text{J}$$

Quick reasonableness check

Energy should increase with both $k$ and $x$, and because $x$ is squared, doubling the stretch would quadruple $E$. The value $\sim 1.7\,\text{J}$ is consistent with a fairly soft spring ($7\,\text{N/m}$) stretched less than 1 meter.