1. Problem 10: Find the acceleration of a block sliding down an inclined plane at 30° with kinetic friction coefficient 0.2. 2. Formula: The acceleration $a$ down the incline is given by $$a = g(\sin\theta - \mu_k \cos\theta)$$ where $g = 9.8$ m/s², $\theta = 30^\circ$, and $\mu_k = 0.2$. 3. Calculate $\sin 30^\circ = 0.5$ and $\cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866$. 4. Substitute values: $$a = 9.8 (0.5 - 0.2 \times 0.866) = 9.8 (0.5 - 0.1732) = 9.8 \times 0.3268 = 3.2 \text{ m/s}^2$$ 5. Problem 11: Find the coefficient of sliding friction $\mu$ given mass 10 kg, incline 30°, and least force up the plane 100 N. 6. The least force $F$ to move the block up is: $$F = mg \sin\theta + \mu mg \cos\theta$$ Rearranged to find $\mu$: $$\mu = \frac{F - mg \sin\theta}{mg \cos\theta}$$ 7. Calculate $mg = 10 \times 9.8 = 98$ N. 8. Substitute values: $$\mu = \frac{100 - 98 \times 0.5}{98 \times 0.866} = \frac{100 - 49}{84.868} = \frac{51}{84.868} = 0.6$$ 9. Problem 12: Find average force on a 0.5 kg ball hitting a wall at 12 m/s and bouncing back at 8 m/s over 0.1 s. 10. Change in velocity $\Delta v = v_{final} - v_{initial} = -8 - 12 = -20$ m/s (negative because direction reverses). 11. Impulse-momentum theorem: $$F_{avg} = m \frac{\Delta v}{\Delta t} = 0.5 \times \frac{-20}{0.1} = -100 \text{ N}$$ 12. The magnitude of average force is 100 N directed opposite to initial velocity. Final answers: - Acceleration down incline: $3.2$ m/s² - Coefficient of friction: $0.6$ - Average force on ball: $100$ N