A block A of mass 4 kg can slide on a rough horizontal table - Leaving Cert Applied Maths - Question 4 - 2014

Now, we need to find the tension T in the string. Using the relationship derived from the acceleration, we substitute back:

Using the earlier equation where:

$T = 8g - 8a$

Substituting the value of a:

$T = 8(9.81) - 8(4)$

Calculating this gives:

$T = 140.96 \, N$

To analyze the forces acting on both the wedge and the particle, we can draw separate free-body diagrams.

1. For the Wedge (mass 8 kg):

   • The forces include: the weight acting downwards (8g), normal force exerted by the floor, and the horizontal reaction from the particle.

2. For the Particle (mass 3 kg):

   • The components of the weight acting downwards, the normal force from the wedge, and the friction force acting parallel to the slope.

These forces can be summarized as:

• On the particle:
  • Downward force: (3g)
  • Normal force at angle 30°
  • Frictional force acting along the slope.

From the equations of motion, we derive:

(3g \sin(30) - f = 3a) (3g \cos(30) - R = 3f / \sin(30))

Using the equations derived:

1. Determine t:

   • The wedge moves (\frac{\sqrt{3}}{3}) cm in t seconds: (a = \frac{22}{35} m/s²)
   • This gives: $t = \frac{\sqrt{3}}{100} + \frac{1}{2} \frac{\sqrt{3}}{35} t^{2}$

2. Calculate x:

   • This distance can be derived using the relationships and substituting: $x = \frac{1}{2} \cdot a \cdot t^2$
   • Hence, we find the values of t and x after plugging the results back into the equations: $x = 0 + \frac{1}{2} \cdot \frac{22}{35} \cdot \frac{\sqrt{3}}{10}$
   • Finally determining: $x \approx 0.22 \, m$