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

To find the coefficient of friction, we start by applying Newton's second law to both masses. For particle B, we have:

1. The weight of B is contributing to the acceleration, hence:

   $2g - T = 2a$

2. For block A on the slope:

   $T - W \, ext{sin} \theta - \mu R = ma$

   Here, W is the weight of block A, $W = mg = 0.2 \times 9.81$ and $R = W \cos \theta$. The perpendicular component of weight can be calculated as:

   $W \cos \theta = 0.2g \cos \theta$

   where $heta$ can be derived as a function of its tangent.

Then, substituting the known values, both equations must be solved simultaneously:

• From particle B's equation:

$T = 2g - 2a$

Substituting $g$ and $a$ gives us:

$T = 2(9.81) - 2 \left(\frac{543}{625} \times 9.81\right)$

Using these two equations, we can solve for eta and thus find the coefficient of friction, resulting in a final value of eta = 0.17.