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

The acceleration of P can be found using the formula:

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

Where:

• (v \approx 3.71 , m/s) (from part a)
• (u = 2 , m/s)
• (t = 3.5 , s)

Substituting these values:

$a = \frac{3.71 - 2}{3.5}\approx 0.490 \, m/s^2$

Thus, the acceleration of P is approximately 0.49 m/s².

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

1. The normal reaction (N) can be calculated as: $N = 0.6 \times g \times \cos(25°)$ Assuming (g \approx 9.81 , m/s^2): $N \approx 0.6 \times 9.81 \times \cos(25°) \approx 0.6 \times 9.81 \times 0.9063 \approx 5.329 \, N$

2. Resolve the forces parallel to the slope: We have: $0.6 \times g \sin(25°) - \mu N = 0.6 \times a$ Substituting values: $0.6 \times 9.81 \times 0.4226 - \mu \cdot 5.329 = 0.6 \times 0.49$ Rearranging gives us: $\mu = \frac{0.6 \times 9.81 \times 0.4226 - 0.6 \times 0.49}{5.329}$ Substituting the calculated values: $\mu \approx 0.41 \text{ or } 0.411$

Thus, the coefficient of friction between P and the plane is approximately 0.41 or 0.411.