Two bodies of masses m_1 = 1 kg and m_2 = 5 kg are connected…
Two bodies of masses $m_1 = 1\ \mathrm{kg}$ and $m_2 = 5\ \mathrm{kg}$ are connected by a light inextensible rope as shown in the diagram; one end of the rope is fixed. The body $m_1$ slides down a frictionless incline of angle $\alpha = 30^\circ$. Find the acceleration $a_2$ of the body $m_2$. Neglect friction and the mass of the pulley.
Explanation of Method
Two bodies on a rope over an incline.
Step 1: The diagram is missing
The statement says one end of the rope is fixed, which usually means a movable pulley, where the two accelerations differ by a factor of 2. Without the diagram, the arrangement cannot be reconstructed with certainty. The original solution treated it as a simple rope over a fixed pulley at the top of the incline, with $m_2$ hanging; that case is worked out here.
Step 2: Simple pulley arrangement
Take $m_2$ moving down as positive: $m_2 g - T = m_2 a$.
\[ T - m_1 g\sin\alpha = m_1 a. \]
\[ a = \frac{m_2 - m_1\sin\alpha}{m_1 + m_2}\,g = \frac{5 - 0.5}{6} \cdot 9.81. \]
\[\begin{gathered} a_2 \approx 7.36\ \mathrm{m/s^2} \\ (m_2 \text{ down, } m_1 \text{ up the incline}) \end{gathered}\]
Rope tension: $T = m_2(g - a) \approx 12.3\ \mathrm{N}$.
Result
\[\boxed{\,\begin{gathered} a_2 \approx 7.36\ \mathrm{m/s^2} \\ (m_2 \text{ down, } m_1 \text{ up the incline}) \end{gathered}\,}\]
Remark
About the original solution. Its numbers are right for this arrangement, but they contradict the statement: with a hanging 5 kg body, $m_1$ is pulled up the incline, not down. So the real figure must be different, most likely with $m_2$ on a horizontal surface or on a movable pulley. The answer should be redone once the figure is available.