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

For mass B (0.8 kg), using Newton's second law: $F_{net} = mg - T$ Where $m = 0.8 ext{ kg}$ and $g = 9.8 ext{ m/s}^2$ (approximated to 0.8): $0.8g - T = 0.8a$ Substituting known values gives: $0.8 \times 9.8 - T = 0.8 \times 3.2$ Thus: $T = 0.8 \times 9.8 - 0.8 \times 3.2 = 5.28 ext{ N or } 5.3 ext{ N}$

For block A (0.5 kg), the forces acting are:

• The weight: $0.5g$ downward
• The normal force: $N$ upward The frictional force is given by: $F_{friction} = \mu N$ Using the equation for block A: $N - T = 0.5a$ Substituting $N = 0.5g$ gives: $0.5g - T = 0.5 \times 3.2$ Substituting $T$: $0.5 \times 9.8 - T = 0.5 \times 3.2$ After calculations, we find: $\mu = 0.75\text{ or } 0.751$

The concept of inextensibility of the string implies that both A and B will have the same acceleration throughout the motion. This means that any change in the position of B directly translates to a corresponding change in A, ensuring that the tensions and forces applied along the string remain consistent throughout the system.