1. **Stating the problem:** Given a mass $m=5.35$ kg, a kinetic friction coefficient $\mu_k=0.433$, and a position-dependent force $$\vec{F}(x) = \left(21.85 \frac{N}{m} x - 9.08 \frac{N}{m^2} x^2\right) \hat{i},$$ with initial position $\vec{r}_i = 0.97\,m \hat{i}$ and initial velocity $\vec{v}_i = 7.91 \frac{m}{s} \hat{i}$. 2. **Understanding the force:** The force depends on position $x$ along the $\hat{i}$ direction. It is a polynomial force with a linear and quadratic term. 3. **Goal:** We can analyze the motion of the mass under this force, including finding acceleration, velocity, or position as functions of time or displacement. 4. **Formula used:** Newton's second law: $$m \vec{a} = \vec{F}_{net}$$ where $\vec{a}$ is acceleration. 5. **Friction force:** The kinetic friction force magnitude is $$F_{fric} = \mu_k m g,$$ acting opposite to the direction of motion. 6. **Net force:** Assuming friction acts opposite to velocity, the net force along $x$ is $$F_{net}(x) = F(x) - F_{fric} \cdot \text{sign}(v).$$ 7. **Acceleration:** $$a(x) = \frac{F_{net}(x)}{m} = \frac{21.85 x - 9.08 x^2 - \mu_k m g \cdot \text{sign}(v)}{m}.$$ 8. **Substitute values:** Given $g=9.8 \frac{m}{s^2}$, $$F_{fric} = 0.433 \times 5.35 \times 9.8 = 22.7\,N.$$ 9. **Final acceleration expression:** $$a(x) = \frac{21.85 x - 9.08 x^2 - 22.7 \cdot \text{sign}(v)}{5.35}.$$ 10. **Interpretation:** This acceleration depends on position $x$ and velocity direction. To solve for motion, one would typically use numerical methods or energy methods. **Summary:** We have expressed the net force and acceleration as functions of position and velocity direction, incorporating friction and the given force.