A particle P of mass 6 kg lies on the surface of a smooth plane - Edexcel - A-Level Maths Mechanics - Question 4 - 2008 - Paper 1

The normal reaction $R$ between the particle P and the plane is derived from considering the vertical forces acting on the particle. The equation can be expressed as:

$R = 6g \cos 30° + 49 \sin θ.$

Calculating $R$, we first find:

Using $g = 9.8 \, ext{m/s}^2$ and plugging in the values:

$R = 6 \times 9.8 \times \frac{\sqrt{3}}{2} + 49 \sin θ.$

To summarize:

$R = 90 \, ext{N} \, \text{or} \, 91 \, ext{N}.$

When the force of 49 N is applied horizontally to P, the forces along the slope of the plane can be analyzed. The net force in the direction of motion is given by:

$49 \cos 30° - 6g \sin 30° = 6a.$

Substituting the known values, we find:

$49 \times \frac{\sqrt{3}}{2} - 6 \times 9.8 \times 0.5 = 6a.$

Calculating this gives:

$\Rightarrow 42.44 - 29.4 = 6a,$

leading to:

$a = \frac{13.04}{6} \approx 2.17 \, \text{m/s}^2.$

Thus, the initial acceleration of P is approximately $2.17 \, \text{m/s}^2$.