Q 11-08-017NEETJEE MainMedium
A load of $200$ N stretches a wire by $2$ mm, within the elastic limit. The elastic potential energy stored in the wire is
Answer: (A) $0.2$ J
Concept. The stretching force grows linearly from $0$ to $F$, so the work done (energy stored) is $U = \dfrac12F\Delta L$.
Step 1. Substitute.
$$U = \frac12 \times 200 \times 2 \times 10^{-3} = 0.2\ \text{J}$$
Answer. $0.2$ J (Option A).
Solution by Sreeraj P, M.Sc Physics