Calculate Velocity of Released Mass Striking Ground in Pulley System

To find the velocity at which the $6\text{ kg}$ mass strikes the ground, we can use either Newton's laws of motion or the principle of conservation of mechanical energy.

Method 1: Using Work-Energy Theorem / Conservation of Energy

Let:

• Mass of the falling block, $m_1 = 6\text{ kg}$
• Mass of the rising block, $m_2 = 2\text{ kg}$
• Height of release, $h = 6\text{ m}$
• Acceleration due to gravity, $g = 10\text{ m/s}^2$

When the $6\text{ kg}$ mass falls by a height $h = 6\text{ m}$, the $2\text{ kg}$ mass rises by the same height $h = 6\text{ m}$. Since the string is inextensible, both blocks move with the same magnitude of velocity $v$.

By conservation of mechanical energy: $\text{Loss in Potential Energy} = \text{Gain in Kinetic Energy}$

$\Delta U_{loss} = m_1 g h - m_2 g h = (m_1 - m_2) g h$ $\Delta K_{gain} = \frac{1}{2} (m_1 + m_2) v^2$

Equating the loss in potential energy to the gain in kinetic energy: $(m_1 - m_2) g h = \frac{1}{2} (m_1 + m_2) v^2$

Substitute the given values: $(6 - 2) \times 10 \times 6 = \frac{1}{2} (6 + 2) v^2$ $4 \times 10 \times 6 = \frac{1}{2} (8) v^2$ $240 = 4 v^2$ $v^2 = 60$ $v = \sqrt{60} \approx 7.74596\text{ m/s} \approx 7.74\text{ m/s}$

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Method 2: Using Newton's Laws and Kinematics

1. Calculate the acceleration ($a$) of the system: $a = \left(\frac{m_1 - m_2}{m_1 + m_2}\right) g$ $a = \left(\frac{6 - 2}{6 + 2}\right) \times 10 = \frac{4}{8} \times 10 = 5\text{ m/s}^2$

2. Use the third equation of motion: $v^2 = u^2 + 2ah$

   Given that the system starts from rest ($u = 0$) and travels a distance $h = 6\text{ m}$: $v^2 = 0 + 2 \times 5 \times 6 = 60$ $v = \sqrt{60} \approx 7.74\text{ m/s}$

Thus, the velocity at which the $6\text{ kg}$ mass strikes the ground is $7.74\text{ m/s}$.