Question

If the system shown in the figure is released from rest, find the speed of the \(6\,\text{kg}\) block just before hitting the ground. \((g = 10\,\text{m/s}^2)\)

• A. \(6.20\,\text{m/s}\)
• B. \(7.74\,\text{m/s}\)
• C. \(4.70\,\text{m/s}\)
• D. \(5.20\,\text{m/s}\)

Answer 1

The Correct Option is B

Concept: This system behaves like an Atwood machine. When two masses are connected by a light string over a frictionless pulley, the acceleration of the system is \[ a = \frac{(m_1 - m_2)g}{m_1 + m_2} \] Once acceleration is known, the velocity can be obtained using the kinematic relation \[ v^2 = u^2 + 2as \] where \begin{itemize} \item \(u\) = initial velocity \item \(a\) = acceleration \item \(s\) = displacement \end{itemize} Step 1: {\color{red}Calculate acceleration of the system.} Given masses: \[ m_1 = 6\,\text{kg}, \qquad m_2 = 2\,\text{kg} \] \[ a = \frac{(6-2)10}{6+2} \] \[ a = \frac{40}{8} = 5\,\text{m/s}^2 \] Step 2: {\color{red}Use the kinematic equation to find final speed.} Initial velocity: \[ u = 0 \] Distance moved by the \(6\,\text{kg}\) block: \[ s = 6\,\text{m} \] Using \[ v^2 = u^2 + 2as \] \[ v^2 = 0 + 2(5)(6) \] \[ v^2 = 60 \] \[ v = \sqrt{60} \] \[ v \approx 7.74\,\text{m/s} \] Thus, the speed of the \(6\,\text{kg}\) block just before hitting the ground is \[ \boxed{7.74\,\text{m/s}} \]