Question

1. two masses, 3 kg and 5 kg, hang on opposite sides of a frictionless pulley. find the acceleration of the system and the tension in the rope.

Explanation:

Step1: Define Variables and Forces

Let \( m_1 = 3\,\text{kg} \), \( m_2 = 5\,\text{kg} \), \( a \) be acceleration, \( T \) be tension. For \( m_2 \) (heavier), net force: \( m_2g - T = m_2a \). For \( m_1 \) (lighter), net force: \( T - m_1g = m_1a \).

Step2: Add Equations to Eliminate \( T \)

Adding \( (m_2g - T) + (T - m_1g) = m_2a + m_1a \). Simplifies to \( (m_2 - m_1)g = (m_1 + m_2)a \).

Step3: Solve for Acceleration \( a \)

Substitute \( m_1 = 3 \), \( m_2 = 5 \), \( g = 9.8\,\text{m/s}^2 \). \( a = \frac{(m_2 - m_1)g}{m_1 + m_2} = \frac{(5 - 3) \times 9.8}{3 + 5} = \frac{19.6}{8} = 2.45\,\text{m/s}^2 \).

Step4: Solve for Tension \( T \)

Use \( T = m_1g + m_1a = m_1(g + a) \). Substitute values: \( T = 3 \times (9.8 + 2.45) = 3 \times 12.25 = 36.75\,\text{N} \).

Answer:

Acceleration: \( 2.45\,\text{m/s}^2 \), Tension: \( 36.75\,\text{N} \)