How much energy is in the elastic potential energy store of a spring with a spring constant of 4 N/m if it is stretched from 1 m to 1.1 m in length?

[tex]E_p[/tex] the elastic potential energy of an ideal spring,
[tex]k[/tex] its spring coefficient, and
[tex]\Delta x[/tex] the distance by which it was stretched or compressed with reference to its initial length.

The question implied that the spring has no elastic potential energy when it was of length 1 meter. 1 meter is therefore its initial length.

The spring has a length of 1.1 meters after the stretch. It has thus been stretched with a displacement [tex]\Delta x[/tex] of [tex]1.1 - 1 = 0.1 \; \text{m}[/tex].

[tex]E_p = 1/2 \; k \cdot \Delta x^2\\\phantom{E_p} = 1/2 \times 4 \; \text{N} \cdot \text{m}^{-1} \times ((1.1 - 1) \; \text{m})^2\\\phantom{E_p} = 0.02 \; \text{N} \cdot \text{m}\\\phantom{E_p} = 0.02 \; \text{J}[/tex]

Make sure that the final expression is in unit Newton [tex]\cdot[/tex] Meters- not cm- which is equivalent to Joules, the unit for energies.