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

To find the acceleration of P, we use the formula: $a = \frac{v - u}{t}$ Substituting the values:

• v = 9.43 m/s (speed at B)
• u = 2 m/s (speed at A)
• t = 3.5 s

Thus, $a = \frac{9.43 - 2}{3.5} = \frac{7.43}{3.5} \approx 2.12 \text{ m/s}^2$. The acceleration of P is approximately 2.12 m/s².

To find the coefficient of friction (μ), we start by calculating the normal reaction (R): $R = mg \cos(25°)$ Substituting the values:

• m = 0.6 kg
• g = 9.81 m/s²

Thus, $R = 0.6 \times 9.81 \cos(25°) \approx 5.29 ext{ N}$

Now, resolving forces parallel to the slope: [ 0.6g \sin(25°) - , \mu \cdot R = 0.6 \cdot a ]

Substituting known values: $0.6 \cdot 9.81 \sin(25°) - \mu \cdot 5.29 = 0.6 \cdot 2.12$ Calculate the left side: $0.6 \cdot 9.81 \sin(25°) \approx 1.47$ Thus, $1.47 - \mu \cdot 5.29 = 1.27$ Rearranging gives: $\mu \cdot 5.29 = 1.47 - 1.27 \approx 0.2$ $\mu = \frac{0.2}{5.29} \approx 0.0378$ After calculating using all given conditions the coefficient of friction μ is approximately 0.0378.