Bodies of masses m_1 = 3 kg and m_2 = 2 kg lie on a smooth…

Bodies of masses $m_1 = 3\ \mathrm{kg}$ and $m_2 = 2\ \mathrm{kg}$ lie on a smooth incline of angle $\alpha = 30^\circ$, joined by a spring of stiffness $k = 1000\ \mathrm{N/m}$. A rope from $m_2$ passes over a pulley at the top to a hanging body $m_3 = 9\ \mathrm{kg}$. Find the extension of the spring while the system moves.

Explanation of Method

Spring between two bodies pulled up an incline.

Step 1: Acceleration

Once the motion is steady, all three bodies share the acceleration $a$: $a = \frac{m_3 - (m_1 + m_2)\sin\alpha}{m_1 + m_2 + m_3}\,g$.

\[ a = \frac{9 - 2.5}{14} \cdot 9.81 \approx 4.55\ \mathrm{m/s^2}. \]

Step 2: Spring force

The spring is the only thing pulling $m_1$ up the slope: $kx - m_1 g\sin\alpha = m_1 a$.

\[ kx = 3\,(4.905 + 4.55) \approx 28.4\ \mathrm{N}. \]

\[x \approx 0.0284\ \mathrm{m} = 2.84\ \mathrm{cm}\]

$30^\circ$ $m_1$ $m_2$ $m_3$ $k$ $a = 4.55\ \mathrm{m/s^2}$ $kx = m_1\,(a + g\sin\alpha) \Rightarrow x \approx 2.84\ \mathrm{cm}$

The hanging 9 kg body pulls $m_2$ and, through the spring, $m_1$ up the smooth incline. The spring alone accelerates $m_1$, so its stretch depends only on what $m_1$ needs.

Result

\[\boxed{\,x \approx 0.0284\ \mathrm{m} = 2.84\ \mathrm{cm}\,}\]

Remark

Check. The original result is correct. In closed form, $kx = \dfrac{m_1 m_3\,g\,(1 + \sin\alpha)}{m_1 + m_2 + m_3}$. (The statement's angle symbol "x" clashes with the extension $x$; here the angle is called $\alpha$.)