Question: PART B Physics 26 A spring is initially unstretched. A force $F$ is used to stretch the spring. The extension $x$ and the energy $E$ stored in the stretched spring are measured for different values of $F$. The graph shows how the energy $E$, in J, varies with the extension squared, $x^2$, in cm$^2$. $E / J$ 0.015 0 0 25 $x^2 / cm^2$ What is the magnitude of $F$ when the spring stores 0.015 J of energy? A 0.30 N B 0.60 N C 1.2 N D 1.5 N E 2.4 N F 3.0 N G 30 N H 60 N Graph: A straight-line graph in the center shows $E$ versus $x^2$, starting at the origin (bottom-left) and rising to the point $(25, 0.015)$ at the top-right; dashed guide lines mark $x^2 = 25$ and $E = 0.015$.

1. **State the problem:** We need to find the magnitude of the force $F$ when the spring stores $0.015$ J of energy. 2. **Recall the formula for energy stored in a spring:** $$E = \frac{1}{2}kx^2$$ where $k$ is the spring constant and $x$ is the extension. 3. **Given graph information:** The graph plots $E$ against $x^2$ and passes through the origin and the point $(25, 0.015)$. 4. **Find the spring constant $k$:** From the formula, $$E = \frac{1}{2}kx^2 \implies k = \frac{2E}{x^2}$$ Using the point $(x^2 = 25, E = 0.015)$, $$k = \frac{2 \times 0.015}{25} = \frac{0.03}{25} = 0.0012 \text{ N/cm}$$ 5. **Convert units of $k$ to N/m:** Since $1 \text{ m} = 100 \text{ cm}$, $$k = 0.0012 \text{ N/cm} = 0.0012 \times 100 = 0.12 \text{ N/m}$$ 6. **Find the extension $x$ when $E = 0.015$ J:** From the graph, $x^2 = 25$ cm$^2$, so $$x = \sqrt{25} = 5 \text{ cm} = 0.05 \text{ m}$$ 7. **Calculate the force $F$ using Hooke's law:** $$F = kx = 0.12 \times 0.05 = 0.006 \text{ N}$$ 8. **Check the options:** The calculated force $0.006$ N is much smaller than all options, so re-examine the unit conversion. 9. **Re-examining step 5:** The spring constant $k$ should be in N/m, but the energy formula uses $x$ in meters squared. Since $x^2$ is given in cm$^2$, convert $x^2$ to m$^2$: $$25 \text{ cm}^2 = 25 \times (0.01)^2 = 25 \times 0.0001 = 0.0025 \text{ m}^2$$ 10. **Recalculate $k$ with $x^2$ in m$^2$:** $$k = \frac{2E}{x^2} = \frac{2 \times 0.015}{0.0025} = \frac{0.03}{0.0025} = 12 \text{ N/m}$$ 11. **Calculate $F$ again:** $$x = \sqrt{0.0025} = 0.05 \text{ m}$$ $$F = kx = 12 \times 0.05 = 0.6 \text{ N}$$ 12. **Final answer:** The magnitude of $F$ is $0.6$ N, which corresponds to option B. **Answer: B 0.60 N**