A light cord passes over a friction-free pulley. Each end of the cord has a mass attached, one mass of 25 kg, and the other 12 kg. Calculate: 4.2.1 The acceleration of the masses. 4.2.2 The tension in the cord.

QUESTION 4.2: DYNAMICS

A light cord passes over a friction-free pulley. Each end of the cord has a mass attached, one mass of $25 \text{ kg}$, and the other $12 \text{ kg}$. Given:

• Mass 1: $m_1 = 25 \text{ kg}$
• Mass 2: $m_2 = 12 \text{ kg}$
• Acceleration due to gravity: $g = 9.81 m/s^2$

4.2.1 The acceleration of the masses.

Step 1: Apply Newton's Second Law to each mass. Let $T$ be the tension in the cord and $a$ be the acceleration of the system. Since $m_1 > m_2$, $m_1$ will accelerate downwards and $m_2$ will accelerate upwards.

For $m_1$ (downwards positive): $m_1 g - T = m_1 a \quad (1)$ For $m_2$ (upwards positive): $T - m_2 g = m_2 a \quad (2)$

Step 2: Solve the system of equations for acceleration ($a$). Add equation (1) and equation (2): $(m_1 g - T) + (T - m_2 g) = m_1 a + m_2 a$ $m_1 g - m_2 g = (m_1 + m_2) a$ Factor out $g$ on the left and $a$ on the right: $g(m_1 - m_2) = a(m_1 + m_2)$ Solve for $a$: $a = \frac{g(m_1 - m_2)}{m_1 + m_2}$ Substitute the given values: $a = \frac{9.81 m/s^2 (25 kg - 12 kg)}{25 kg + 12 kg}$ $a = \frac{9.81 m/s^2 (13 kg)}{37 kg}$ $a = \frac{127.53}{37} m/s^2$ $a \approx 3.446756 m/s^2$ Rounding to three significant figures: $\boxed{3.45 m/s^2}$

4.2.2 The tension in the cord.

Step 3: Substitute the calculated acceleration into one of the original equations to find the tension ($T$). Using equation (2): $T - m_2 g = m_2 a$ $T = m_2 a + m_2 g$ $T = m_2 (a + g)$ Substitute the values: $T = 12 kg (3.446756 m/s^2 + 9.81 m/s^2)$ $T = 12 kg (13.256756 m/s^2)$ $T \approx 159.081 N$ Rounding to three significant figures: $\boxed{159 N}$