1. **Problem Statement:** We have a block of mass $m=3.66$ kg subjected to a net force given by $$\vec{F}_{net}(x) = \left(-3.69 \frac{N}{m} x - 2.26 \frac{N}{m^3} x^3 \right) \hat{i}$$ with initial position $x=0$ m and initial velocity $\vec{v} = -4.42 \frac{m}{s} \hat{i}$. We want to find: - The smallest value of $x$ the block reaches (turning point). - The block's speed when it reaches $x=1.28$ m. 2. **Relevant Formulas and Concepts:** - The force depends on position, so we can find the potential energy $U(x)$ by integrating the force: $$F_x = -\frac{dU}{dx} \implies U(x) = -\int F_x \, dx$$ - Total mechanical energy $E$ is conserved: $$E = K + U = \text{constant}$$ where kinetic energy $K = \frac{1}{2} m v^2$. - At turning points, velocity $v=0$ and all energy is potential. 3. **Find the potential energy function $U(x)$:** Given $$F_x = -3.69 x - 2.26 x^3$$ Then $$U(x) = -\int F_x \, dx = -\int (-3.69 x - 2.26 x^3) \, dx = \int (3.69 x + 2.26 x^3) \, dx$$ Calculate the integral: $$U(x) = 3.69 \frac{x^2}{2} + 2.26 \frac{x^4}{4} + C = 1.845 x^2 + 0.565 x^4 + C$$ Choose $U(0)=0$ so $C=0$. 4. **Calculate total energy $E$ at initial position $x=0$:** At $x=0$, potential energy $U(0)=0$. Initial velocity $v_0 = -4.42$ m/s, so kinetic energy: $$K_0 = \frac{1}{2} m v_0^2 = \frac{1}{2} \times 3.66 \times (4.42)^2 = 35.7 \text{ J (approx)}$$ Total energy: $$E = K_0 + U(0) = 35.7 \text{ J}$$ 5. **Find smallest $x$ (turning point) where velocity is zero:** At turning point, kinetic energy is zero, so $$E = U(x) = 1.845 x^2 + 0.565 x^4$$ Set equal to total energy: $$1.845 x^2 + 0.565 x^4 = 35.7$$ Divide both sides by 0.565: $$\frac{1.845}{0.565} x^2 + x^4 = \frac{35.7}{0.565}$$ $$3.2637 x^2 + x^4 = 63.19$$ Let $y = x^2$, then $$y^2 + 3.2637 y - 63.19 = 0$$ Solve quadratic for $y$: $$y = \frac{-3.2637 \pm \sqrt{(3.2637)^2 + 4 \times 63.19}}{2}$$ Calculate discriminant: $$D = 10.65 + 252.76 = 263.41$$ $$\sqrt{D} = 16.23$$ Positive root: $$y = \frac{-3.2637 + 16.23}{2} = \frac{12.9663}{2} = 6.483$$ Negative root is discarded since $y=x^2 \geq 0$. So $$x = \pm \sqrt{6.483} = \pm 2.55 \text{ m}$$ The smallest value reached is $x = -2.55$ m. 6. **Find speed at $x=1.28$ m:** Use energy conservation: $$E = K + U$$ $$K = E - U(1.28)$$ Calculate $U(1.28)$: $$U(1.28) = 1.845 (1.28)^2 + 0.565 (1.28)^4$$ Calculate powers: $$(1.28)^2 = 1.6384$$ $$(1.28)^4 = (1.6384)^2 = 2.683$$ Calculate $U(1.28)$: $$U(1.28) = 1.845 \times 1.6384 + 0.565 \times 2.683 = 3.025 + 1.516 = 4.541 \text{ J}$$ Kinetic energy at $x=1.28$: $$K = 35.7 - 4.541 = 31.16 \text{ J}$$ Calculate speed: $$K = \frac{1}{2} m v^2 \implies v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2 \times 31.16}{3.66}} = \sqrt{17.02} = 4.13 \frac{m}{s}$$ **Final answers:** - Smallest $x$ reached: $\boxed{-2.55 \text{ m}}$ - Speed at $x=1.28$ m: $\boxed{4.13 \frac{m}{s}}$