A box of mass 30 kg is being pulled along rough horizontal ground at a constant speed using a rope - Edexcel - A-Level Maths Mechanics - Question 6 - 2007 - Paper 1

To find the value of P, we begin by analyzing the forces acting on the box. We know there is a frictional force at play, described by the equation:
$F_{friction} = ext{Coefficient of friction} imes R$
where $R$ is the normal reaction force.

Since the box moves at a constant speed, the horizontal forces must balance, leading to:
$P imes ext{cos}(20°) = ext{frictional force}$

In the vertical direction, we have:
$R + P imes ext{sin}(20°) = 30g$

Substituting the known values:

1. Rearranging gives us:
   $R = 30g - P imes ext{sin}(20°)$
2. We can express the frictional force in terms of $P$:
   $P imes ext{cos}(20°) = 0.4R$
3. Substitute for $R$:
   $P imes ext{cos}(20°) = 0.4(30g - P imes ext{sin}(20°))$
4. Solve for P:
   Substituting $g ≈ 9.81 ext{ m/s}^2$ gives us an equation in terms of $P$. After working through the algebra, we find:
   $P ≈ 110 ext{ N}$.