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 1

We can find the acceleration using:

$a = \frac{v - u}{t}$

Where:

• $v \approx 3.71$ m/s
• $u = 2$ m/s
• $t = 3.5$ s

Calculating:

$a = \frac{3.71 - 2}{3.5} = \frac{1.71}{3.5} \approx 0.49 \text{ m/s}^2$.

To find the coefficient of friction, we first need to calculate the normal reaction:

$R = 0.6g \cos 25°$

The weight component along the slope:

$0.6g \sin 25° - \mu R = 0.6 \cdot a$

Substituting values, where $g \approx 9.8$ m/s²:

1. $R = 0.6 \cdot 9.8 \cdot \cos(25°)$
2. Resolve parallel to the slope:

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

Solving for $\mu$ yields: $\mu \approx 0.41 ext{ or } 0.411$.