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 solve for the coefficient of friction, we start by applying Newton's second law to both block A and block B.

For block B, moving downward:

$$2g - T = 2a$$

Here, g = 9.81 m/s² and a = \frac{543}{625}g. This translates to:

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

Now, we rearrange to find T:

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

For block A, at rest on the inclined plane, we consider the forces:

$$T - W_{||} = \mu R$$

Where W_{||} is the parallel component of gravity, expressibly as:

$$W_{||} = mg \sin(θ) = 0.2g\left(\frac{7}{25}\right)$$

The normal force R can be defined as:

$$R = mg \cos(θ) = 0.2g\left(\frac{24}{25}\right)$$

Substituting into the friction equation yields:

$$T - 0.2g\left(\frac{7}{25}\right) = \mu \cdot 0.2g\left(\frac{24}{25}\right)$$

From the previous steps' T, we substitute it back into the equation:

Now equate the two equations for T:

$$2g - 2\left( \frac{543}{625}g \right) - 0.2g\left(\frac{7}{25}\right) = \mu \cdot 0.2g\left(\frac{24}{25}\right)$$

Simplifying this will yield μ = 0.17 when all constants and expressions are evaluated.