Block A, of mass 0.2 kg, lies at rest on a rough plane - AQA - A-Level Maths Mechanics - Question 18 - 2020 - Paper 2

To analyze the forces acting on block A and B, we set up the equations based on Newton's laws.

1. For block B: The downward force due to gravity is given by: $F_B = mg = 2 imes 9.81 = 19.62 ext{ N}$ The tension in the string is denoted by T. Using Newton's second law for block B moving downwards: $2g - T = 2a$ where acceleration, $a = \frac{543}{625} ext{ ms}^{-2}$. Thus, $2g - T = 2 \left(\frac{543}{625}\right)$ Simplifying gives us: $T = 19.62 - \frac{1086}{625} = 19.62 - 1.7376 = 17.8824 ext{ N}$

2. For block A: The forces acting along the incline include the component of weight and the tension from the string. The weight component down the incline is: $W_{A} = mg \sin(\theta)$ where $W_{A} = 0.2 imes 9.81 \times \sin(\theta)$, and the frictional force is given by: $F_{friction} = \\mu imes N$ where $N = mg \cos(\theta)$. Using these equations, we have: $T - W_{A} - F_{friction} = ma$ Substituting: $T - (0.2 imes 9.81 \sin(\theta)) - (\mu \times (0.2 imes 9.81 \cos(\theta))) = (0.2 \times a)$

3. Now substituting the values: $T - (0.2 \times 9.81 \times \frac{7}{25}) - (\mu \times (0.2 \times 9.81 \times \frac{24}{25})) = 0.2 \times \frac{543}{625}$ Rearranging this to solve for \mu gives: $\mu = 0.17$