1. **Problem:** A spring requires a 10 N force to stretch it 3 cm. Find the work to increase the stretch of the spring from 5 cm to 7 cm. 2. **Formula and Explanation:** The force required to stretch or compress a spring is given by Hooke's Law: $$F(x) = kx$$ where $k$ is the spring constant and $x$ is the displacement from the natural length. Work done stretching the spring from $a$ to $b$ is: $$W = \int_a^b F(x) \, dx = \int_a^b kx \, dx$$ 3. **Find the spring constant $k$:** Given $F = 10$ N at $x = 3$ cm = 0.03 m, $$k = \frac{F}{x} = \frac{10}{0.03} = 333.33 \text{ N/m}$$ 4. **Calculate work from 5 cm to 7 cm:** Convert cm to meters: 5 cm = 0.05 m, 7 cm = 0.07 m. $$W = \int_{0.05}^{0.07} 333.33x \, dx$$ 5. **Integrate:** $$W = 333.33 \int_{0.05}^{0.07} x \, dx = 333.33 \left[ \frac{x^2}{2} \right]_{0.05}^{0.07}$$ 6. **Evaluate:** $$W = 333.33 \times \frac{1}{2} (0.07^2 - 0.05^2) = 166.665 (0.0049 - 0.0025) = 166.665 \times 0.0024 = 0.4 \text{ J}$$ **Final answer:** The work required to stretch the spring from 5 cm to 7 cm is approximately $0.4$ joules.