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

The frictional force can be expressed in terms of the normal reaction force (R) and the coefficient of friction (( \mu = 0.5 )). Thus:

F = 0.5 R$$

In the vertical direction, the sum of vertical forces must equal zero because the box moves at constant speed. Thus, we have:

$mg + 100\sin(30°) = R$

Substituting ( \sin(30°) = 0.5 ), gives:

$mg + 100 \times 0.5 = R$ ( R = mg + 50 )

Substituting ( F = 86.6 ) N into ( F = 0.5 R ):

$86.6 = 0.5(mg + 50)$

We can assume ( g = 9.8 , \text{m/s}^2 ):

Expanding the equation:

$173.2 = mg + 50$

This simplifies to:

$mg = 173.2 - 50 \Rightarrow mg = 123.2$

Finally, resolving for m:

$m = \frac{123.2}{9.8} \approx 12.57 \, \text{kg}$

Thus, the mass of the box is approximately 12.6 kg.