4. (a) Two particles of masses 2 kg and 3 kg are connected by a taut, light, inextensible string which passes over a smooth light pulley - Leaving Cert Applied Maths - Question 4 - 2012

To find the tension, we can substitute the value of $a$ back into one of the equations. Using the equation for the 2 kg mass:

$T - 2g = 2a$

Substituting $a = 2$ and $g = 10$:

$T - 20 = 4$

Thus, we find:

$T = 24 \: \text{N}$

For the 9 kg mass on a rough horizontal plane:

1. Weight ($9g$) acting downwards.
2. Normal force ($R$) acting upwards.
3. Frictional force (rac{1}{3}R) acting to the left.
4. Tension ($T$) acting to the right.

For the 12 kg mass on a smooth plane:

1. Weight ($12g$) acting downwards.
2. Normal force ($R$) acting perpendicular to the plane.
3. Tension ($T$) acting upwards along the string.

Setting up the equations based on both masses: For the 12 kg mass:

$12g ext{cos}(30^ ext{o}) - T = 12a$

For the 9 kg mass:

T - rac{1}{3}R = 9a

Using the relationship, we find the common acceleration by solving these equations. Substituting the known values gives:

$6g - 3g = 21a$

So,

$a = \frac{10}{7} \: \text{m/s}^2 ext{ which is approximately } 1.43 \: \text{m/s}^2$.

Using the value of $a$ found in the second step, we substitute back into the equation for tension derived from the motion of the 12 kg mass:

$T = 3g + 9a$

Substituting $g = 10 \: \text{m/s}^2$ and $a = \frac{10}{7} ext{ gives } T = \frac{300}{7} ext{ N} \text{ which yields approximately } 42.9 ext{ N}.$