4. (a) Two particles of masses 9 kg and 5 kg are connected by a taut, light, inextensible string which passes over a smooth light pulley - Leaving Cert Applied Maths - Question 4 - 2008

Using the value of acceleration found previously, we can now find the tension in the string. Using the equation for the 5 kg mass: $T - 5g = 5a$

Substituting for a:

$T - 5 imes 9.8 = 5 imes 2.80$
Solving: $T - 49 = 14$
Thus: $T = 63 ext{ N}$.

When analyzing the second part, we follow a similar approach: Using the distance formula: $s = ut + \frac{1}{2}at^2$
Given that initial velocity (u) is 0:

$1 = 0 + \frac{1}{2} a(\sqrt{2})^2$
Therefore: $a = 1 ext{ m/s²}$.

Examining the forces acting on the 6 kg mass on the inclined plane: Using: $6g \sin(30^{\circ}) - T = 6a$
Substituting for a: $6g \cdot 0.5 - T = 6(1)$
Thus: $T = 3g - 6 = 24 ext{ N}$.

For the 3 kg mass on the rough surface: Using:
$T - μR = 3a$ Where R is the normal force, thus R = 3g.
Substituting: $T - μ(3g) = 3a$
By substituting values of T and a: $24 - 3μg = 3(1)$
Thus: $24 - 3μ(9.8) = 3$
Solving gives, $21 = 29.4μ$
Therefore,
$$μ = rac{21}{29.4} = rac{7}{10}.$