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

Using Newton's second law for B:

$0.8g - T = 0.8a$

Where:

• $g = 9.8 \, m/s^2$
• $a = 3.2 \, m/s^2$

Thus,

$T = 0.8g - 0.8a$

Substituting in the values:

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

Calculating:

$T = 7.84 - 2.56 = 5.28 \, N \text{ or } 5.3 \, N$

For block A, applying forces:

$F = \mu \cdot (0.5g)$

And from Newton's second law:

$T - F = 0.5a$

Substituting for $F$:

$T - \mu (0.5g) = 0.5 \times 3.2$

Rearranging gives:

$\mu = \frac{T - 0.5 \times 3.2}{0.5g}$

Substituting in our previous result for T:

$\mu = \frac{5.3 - 0.5 \times 3.2}{0.5 \times 9.8}$

Solving gives:

$\mu = \frac{5.3 - 1.6}{4.9} = \frac{3.7}{4.9} \approx 0.7551 \text{ or } 0.75$

The inextensibility of the string implies that any movement of block B directly results in an equal movement of block A. Therefore, when calculating the acceleration and tensions, we assume both blocks accelerate with the same magnitude. This property allows us to connect their motion directly, ensuring that $B$'s descent leads to the same upward movement of $A$.