Elastic Potential Energy Of A Spring

Elastic potential energy stored in an ideal spring obeying Hooke's law is $U=\frac{1}{2}k{x}^2$, where $k$ is the force constant and $x$ is the deformation from the natural length. The energy depends on the square of the displacement, so compression and extension of equal magnitude store identical energy. Inserting $k=200\text{ N/m}$ and $x=0.1\text{ m}$ gives $U=\frac{1}{2}(200)(0.1{)}^2=\frac{1}{2}(200)(0.01)=1\text{ J}$. The option 2 J omits the one-half factor. The option 10 J wrongly uses $x$ linearly instead of $x^2$. The option 0.5 J drops the factor relating force constant to energy. Because the restoring force grows linearly with displacement, the average force exerted during the compression is exactly half the maximum force, and this averaging is the physical origin of the one-half factor in the energy expression. This reflects the NCERT derivation of spring energy as the area under the force-extension line. As a unit check, $\text{N/m}\times {\text{m}}^2$ equals $\text{N}\cdot \text{m}$, that is joules, confirming both the dimensions and the modest stored value.