A block of wood A of mass 0.5 kg rests on a rough horizontal table and is attached to one end of a light inextensible string - Edexcel - A-Level Maths Mechanics - Question 5 - 2005 - Paper 1

We apply Newton's Second Law for mass B:

$F_{net} = ma$

where the forces acting on B are its weight and the tension in the string:

$0.8g - T = 0.8 \cdot a$

Inserting $g = 9.8 \, \text{m/s}^2$ and $a = 3.2 \, \text{m/s}^2$:

$0.8 \cdot 9.8 - T = 0.8 \cdot 3.2$

Calculating:

$7.84 - T = 2.56 \Rightarrow T = 7.84 - 2.56 = 5.28 \, \text{N}$

Thus, the tension in the string is $5.28 \, \text{N}$.

For block A, we also apply Newton's Second Law:

The forces acting on A are:

• The frictional force $F = μ \cdot N$ where $N = mg = 0.5g$
• The tension in the string T

The equation can be written as:

$T - F = 0.5a$

Substituting for friction:

$T - μ \cdot (0.5 \cdot 9.8) = 0.5 \cdot 3.2$

From part (b), we substitute $T = 5.28$ N:

$5.28 - μ \cdot 4.9 = 1.6$

Solving for μ:

$5.28 - 1.6 = μ \cdot 4.9 \Rightarrow 3.68 = μ \cdot 4.9 \Rightarrow μ = \frac{3.68}{4.9} \approx 0.75$

Thus, the value of μ is approximately $0.75$.

In my calculations, I utilized the information that the string is inextensible to conclude that the accelerations of both objects A and B are equal. This means that when mass B descends by a certain distance, mass A experiences the same acceleration in the horizontal direction due to the tension in the string being constant throughout. Therefore, the understanding of constant acceleration helped in setting up the equations explicitly for both masses.