The velocity at which \(6\,\text{kg}\) mass (shown in figure) strikes the ground when it is released from a height of \(6\,\text{m}\) above the ground is _____ m/s. Assume pulley is massless and string is light and inextensible. (Take \(g=10\,\text{m/s}^2\)).
Step 1: Understanding the Question:
We have a pulley system with masses 6 kg and 2 kg. We need to find the final speed of the 6 kg mass when it travels 6 m downwards from rest. Step 2: Key Formula or Approach:
1. Find the acceleration (\(a\)) of the system: \(a = \frac{(m_1 - m_2)g}{m_1 + m_2}\).
2. Use the kinematic equation: \(v^2 = u^2 + 2as\). Step 3: Detailed Explanation:
1. Given \(m_1 = 6 \text{ kg}\), \(m_2 = 2 \text{ kg}\), \(s = 6 \text{ m}\), and \(u = 0\).
2. Acceleration of the system:
\[ a = \frac{(6 - 2) \times 10}{6 + 2} = \frac{4 \times 10}{8} = 5 \text{ m/s}^{2} \]
3. Final velocity \(v\):
\[ v^2 = 0 + 2 \times 5 \times 6 = 60 \]
\[ v = \sqrt{60} \approx 7.746 \text{ m/s} \] Step 4: Final Answer:
The velocity of striking the ground is 7.74 m/s.
