A particle P of mass 0.6 kg slides with constant acceleration down a line of greatest slope of a rough plane, which is inclined at 25° to the horizontal - Edexcel - A-Level Maths Mechanics - Question 5 - 2013 - Paper 2

From the earlier calculations, we found the acceleration as:

$a = 0.49 \text{ m/s}^2$

To find the coefficient of friction ($\mu$), we first calculate the normal reaction:

$R = mg \cos(25°)$

Where:

• $m = 0.6 \text{ kg}$ (mass)
• $g \approx 9.8 \text{ m/s}^2$ (acceleration due to gravity)

Calculating:

$R = 0.6 \cdot 9.8 \cdot \cos(25°) \approx 0.6 \cdot 9.8 \cdot 0.9063 \approx 5.327$

Now, resolve the forces parallel to the slope:

$mg \sin(25°) - \mu R = ma$

Substituting values gives:

$0.6 \cdot 9.8 \cdot \sin(25°) - \mu \cdot 5.327 = 0.6 \cdot 0.49$

Calculating:

$0.6 \cdot 9.8 \cdot 0.4226 - \mu \cdot 5.327 = 0.294$

Solving for $\mu$ leads to:

$2.478 - \mu \cdot 5.327 = 0.294 \implies \mu \cdot 5.327 = 2.184 \implies \mu = \frac{2.184}{5.327} \approx 0.41$