Change in elastic potential energy equals average tension times change in extension for linear wire

Steps:

• Identify extension $e_1$ at 100 N from the graph.
• Identify extension $e_2$ at 150 N from the graph.
• Compute change in extension: $\Delta e = e_2 - e_1$.
• Compute $\Delta U = \frac{100 + 150}{2} \times \Delta e$ in joules, then convert to mJ.

Why B is correct:

• For Hooke's law, elastic potential energy change is the trapezoidal area under the tension-extension curve, $\frac{T_1 + T_2}{2} \Delta e = 19$ mJ from graph values.

Why the others are wrong:

• A: Uses minimum tension instead of average, underestimating energy.
• C: Uses maximum tension times $\Delta e$, overestimating area.
• D: Likely misreads graph scale, doubling the correct value.

Final answer: B