Question

Two masses \(\mathrm{m_1}\) and \(\mathrm{m_2}\) are attached to a string which passes over a frictionless smooth pulley. When \(\mathrm{m_1} = 10\ \mathrm{kg}\), \(\mathrm{m_2} = 6\ \mathrm{kg}\) the acceleration of masses is

• A. \(20\ \mathrm{m/s^2}\)
• B. \(5\ \mathrm{m/s^2}\)
• C. \(2.5\ \mathrm{m/s^2}\)
• D. \(10\ \mathrm{m/s^2}\)

Hint:

Heavier mass accelerates downward, lighter mass accelerates upward.

Answer 1

The Correct Option is C

To find the acceleration of the masses, we use the concept of Newton's Second Law and analyze the forces acting on both masses \( m_1 \) and \( m_2 \). Let's proceed step-by-step:

The forces acting on \( m_1 \) are its weight \( m_1g \) going downward and the tension \( T \) in the string going upward. \(m_1g - T = m_1a\)

The forces acting on \( m_2 \) are the tension \( T \) going upward and its weight \( m_2g \) going downward. \(T - m_2g = m_2a\)

By adding the two equations above, the tension \( T \) cancels out: \(m_1g - m_2g = m_1a + m_2a\)

Factor out the common terms and solve for \( a \): \(a = \frac{(m_1 - m_2)g}{m_1 + m_2}\)

Substitute the given values \( m_1 = 10 \ \text{kg} \), \( m_2 = 6 \ \text{kg} \), and \( g = 9.8 \ \text{m/s}^2 \): \(a = \frac{(10 - 6) \times 9.8}{10 + 6} = \frac{4 \times 9.8}{16} = \frac{39.2}{16} = 2.45 \ \text{m/s}^2\)

Rounding off the calculated acceleration \( 2.45 \ \text{m/s}^2 \) to a single decimal place gives \( 2.5 \ \text{m/s}^2 \).

Therefore, the correct answer is \(2.5 \ \text{m/s}^2\).