Masses of 1 kg and 4 kg are connected by a taut, light, inextensible string which passes over a smooth light fixed pulley - Leaving Cert Applied Maths - Question 4 - 2016

To find the tension in the string, we can use the equation derived earlier. We know: $T - g = 1a$ Substituting $a = 6 ext{ m/s}^2$ into the equation gives: $T - 10 = 1(6)$ $T = 10 + 6 = 16 ext{ N}$

For mass 8 kg (on rough surface):

1. Weight $W_3 = 8g$ acting downwards.
2. Normal force $N$ acting upwards.
3. Tension $T$ acting to the right.
4. Frictional force f = rac{1}{4}N acting to the left.

For mass 12 kg (on inclined plane):

1. Weight $W_4 = 12g$ acting downwards at an angle to the normal.
2. Normal force acts perpendicular to the inclined surface.
3. Tension $T$ acting upwards along the string.

Applying Newton’s second law for both masses: For 8 kg mass: $12g imes ext{sin} heta - T = 8a$ For 12 kg mass: $T - 8g = 12a$

Let's substitute the value of gravity $g = 10 ext{ m/s}^2$ and $heta$ corresponding to an heta = rac{3}{4} to calculate acceleration. By resolving and simplifying these equations: $T - 60 = 12a$ From the two forces equations, we can find: $a = 1.8 ext{ m/s}^2$

Using the value of acceleration from the previous part, we can now find the tension: Substituting into the equation for the 12 kg mass: $T - 60 = 12(1.8)$ $T = 60 + 21.6 = 81.6 ext{ N}$

Using the equation of motion, where initial velocity $u = 0$, acceleration $a = 1.8 ext{ m/s}^2$, and time $t = 2 ext{ s}$: $v = u + at$ $v = 0 + 1.8 imes 2 = 3.6 ext{ m/s}$