A block of mass $\frac{2}{\sqrt{2}}$ kg rests on a rough plane inclined at 45° to the horizontal - Leaving Cert Applied Maths - Question 4 - 2011

To find the acceleration of the 4 kg mass, we will analyze the forces acting on both the 2 kg block on the incline and the 4 kg mass hanging vertically.

Step 1: Identify the forces acting on the 4 kg mass.

• Weight: $4g$ downwards (where $g = 9.81 \, \text{m/s}^2$)
• Tension: $T$ upwards.

Using Newton's second law, we have: $4g - T = 4a$

Step 2: Identify the forces acting on the 2 kg block on the incline.

• Weight: $2g$ at an angle of 45°
• Normal force: $R$
• Frictional force: $f = \mu R$ (where $\mu = \frac{1}{4}$)

The components of the weight are:

• Parallel: $2g \sin(45°) = \frac{2g}{\sqrt{2}}$
• Perpendicular: $2g \cos(45°) = \frac{2g}{\sqrt{2}}$

Applying Newton's second law along the incline: $R - \frac{2g}{\sqrt{2}} - \mu R = 2a$
This simplifies to: $R(1 + \frac{1}{4}) = \frac{2g}{\sqrt{2}} + 2a$
So, $R = \frac{\frac{2g}{\sqrt{2}} + 2a}{1.25}$

Step 3: Solve the equations simultaneously.

• Substitute $R$ into the friction force calculation, and rearrange both equations to find $a$ (the acceleration of the system).
  After performing the calculations, we find that: $a = 1.6 \, \text{m/s}^2$.