1. **State the problem:** A 130 kg block slides down an inclined plane with a coefficient of kinetic friction $\mu_k = 0.25$. We need to find (a) the net force on the block and (b) the acceleration of the block down the incline. 2. **Identify forces and formulas:** Let the incline angle be $\theta$. Since it is not given, we assume it is known or we express answers in terms of $\theta$. The forces acting on the block are: - Gravitational force component down the incline: $F_g = mg\sin\theta$ - Frictional force opposing motion: $F_f = \mu_k N = \mu_k mg\cos\theta$ Net force down the incline is: $$F_{net} = mg\sin\theta - \mu_k mg\cos\theta$$ Acceleration is given by Newton's second law: $$a = \frac{F_{net}}{m} = g(\sin\theta - \mu_k \cos\theta)$$ 3. **Calculate net force:** $$F_{net} = 130 \times 9.8 \times (\sin\theta - 0.25 \cos\theta) = 1274 (\sin\theta - 0.25 \cos\theta)$$ 4. **Calculate acceleration:** $$a = 9.8 (\sin\theta - 0.25 \cos\theta)$$ **Summary:** - Net force on the block is $F_{net} = 1274 (\sin\theta - 0.25 \cos\theta)$ Newtons. - Acceleration of the block is $a = 9.8 (\sin\theta - 0.25 \cos\theta)$ m/s$^2$. If the incline angle $\theta$ is provided, substitute its value to get numerical answers.