A mass of 4kg rests on a horizontal table(coefficient of friction 1/2) and is connected by a light in extensible string passing over smooth pulley at the age of table carrying a mass 6kg. Find the acceleration of each particle.

For the mass $m_1 = 4 \, \text{kg}$ on the horizontal table:

• The normal force $N$ acts upwards, balancing its weight $W_1 = m_1 g$.
• The tension $T$ acts horizontally, pulling it towards the pulley.
• The kinetic friction force $f_k$ acts horizontally, opposing the motion. For the hanging mass $m_2 = 6 \, \text{kg}$:
• Its weight $W_2 = m_2 g$ acts downwards.
• The tension $T$ acts upwards.

For $m_1$:
In the vertical direction, the mass is in equilibrium, so $N - m_1 g = 0$.

$$N = m_1 g = 4g$$

The kinetic friction force is $f_k = \mu_k N$. Given $\mu_k = \frac{1}{2}$.

$$f_k = \frac{1}{2} (4g) = 2g$$

In the horizontal direction, applying Newton's Second Law ($\sum F = ma$):

$$T - f_k = m_1 a$$ $$T - 2g = 4a \quad (1)$$

For $m_2$:
Applying Newton's Second Law in the vertical direction (assuming $m_2$ moves downwards):

$$m_2 g - T = m_2 a$$ $$6g - T = 6a \quad (2)$$

Add equation (1) and equation (2) to eliminate $T$:

$$(T - 2g) + (6g - T) = 4a + 6a$$ $$4g = 10a$$ $$a = \frac{4g}{10} = \frac{2g}{5}$$

Using the standard value for acceleration due to gravity, $g = 9.8 \, \text{m/s}^2$:

$$a = \frac{2 \times 9.8 \, \text{m/s}^2}{5}$$ $$a = \frac{19.6 \, \text{m/s}^2}{5}$$ $$a = 3.92 \, \text{m/s}^2$$

Since the string is inextensible, both particles will have the same magnitude of acceleration.

The acceleration of each particle is $\text{3.92 m/s}^2$.