A small box is pushed along a floor - Edexcel - A-Level Maths Mechanics - Question 3 - 2010 - Paper 1

Since the box moves with constant speed, the net force in the horizontal direction is zero. Therefore, we can express this relationship as:

$F_x = \mu R$

where ( \mu ) is the coefficient of friction ( \frac{1}{2} ), and ( R ) is the normal reaction force.

Next, we consider the forces in the vertical direction. The normal force can be stated as:

$R = mg + 100 \sin(30^{\circ})$

Substituting ( \sin(30^{\circ}) = \frac{1}{2} ) leads to:

$R = mg + 50$.

We can substitute ( R ) from the vertical forces into the horizontal forces equation:

$50\sqrt{3} = \mu (mg + 50)$

Substituting ( \mu = \frac{1}{2} ):

$50\sqrt{3} = \frac{1}{2} (mg + 50)$

Upon simplifying and solving for ( mg ):

$mg + 50 = 100\sqrt{3} \Rightarrow mg = 100\sqrt{3} - 50$.

Using ( g \approx 9.81 , \text{m/s}^2 ), we have:

$m = \frac{100\sqrt{3} - 50}{9.81} \approx 12.6 \, \text{kg}$.