1. **Problem Statement:** We have a block of mass $m=5.35$ kg subjected to a position-dependent force along the x-axis given by $$\vec{F}(x) = \left(21.85\frac{N}{m}x - 9.08\frac{N}{m^2}x^2\right)\hat{i}$$ with initial position $\vec{r}_i = 0.97\,m\hat{i}$ and initial velocity $\vec{v}_i = 7.91\frac{m}{s}\hat{i}$. The coefficient of kinetic friction is $\mu_k = 0.433$. 2. **Goal:** Analyze the force function and understand the motion of the block under this force and friction. 3. **Force Function Analysis:** The force depends on position $x$ as $$F(x) = 21.85x - 9.08x^2$$ where $F$ is in Newtons and $x$ in meters. 4. **Important Concepts:** - The force is nonlinear due to the $x^2$ term. - The friction force opposing motion is $F_f = \mu_k mg$. - Newton's second law: $F_{net} = ma = m\frac{d^2x}{dt^2}$. 5. **Net Force on the Block:** $$F_{net}(x) = F(x) - F_f = 21.85x - 9.08x^2 - \mu_k mg$$ Substitute values: $$F_f = 0.433 \times 5.35 \times 9.8 = 22.68\,N$$ So, $$F_{net}(x) = 21.85x - 9.08x^2 - 22.68$$ 6. **Equation of Motion:** Using $F_{net} = m a$: $$5.35 \frac{d^2x}{dt^2} = 21.85x - 9.08x^2 - 22.68$$ or $$\frac{d^2x}{dt^2} = \frac{21.85}{5.35}x - \frac{9.08}{5.35}x^2 - \frac{22.68}{5.35}$$ 7. **Interpretation:** This nonlinear second-order differential equation governs the block's motion considering the position-dependent force and friction. 8. **Graph Shape:** The force $F(x)$ is a quadratic function opening downward (since coefficient of $x^2$ is negative), with zeros at $x=0$ and $x=\frac{21.85}{9.08} \approx 2.41$ m. 9. **Summary:** The block experiences a nonlinear restoring force and constant friction opposing motion. The initial conditions $x=0.97$ m and $v=7.91$ m/s set the starting state for solving the motion equation. **Final note:** To fully solve for $x(t)$, numerical methods or further analysis are needed due to nonlinearity.