Question

Two masses as shown in the figure are suspended from a massless pulley. The acceleration of the system when masses are left free is

• A. $\frac{2g}{3}$
• B. $\frac{g}{3}$
• C. $\frac{g}{9}$
• D. $\frac{g}{7}$

Answer 1

The Correct Option is B

 To solve this problem, we need to determine the acceleration of the system of masses when they are left to move freely under gravity. Let's denote the two masses as \(m_1\) and \(m_2\), with \(m_1 < m_2\), such that the system will move downwards on the side of \(m_2\).

We use Newton's Second Law of Motion which states:

\(F = ma\)

For the mass \(m_2\) descending, the force equation is:

\(m_2g - T = m_2a\)

For the mass \(m_1\) ascending, the force equation is:

\(T - m_1g = m_1a\)

Where:

• \(T\) is the tension in the string.
• \(a\) is the acceleration of the masses.

Adding these two equations to eliminate \(T\), we get:

\(m_2g - m_1g = m_2a + m_1a\)

Combine like terms:

\((m_2 - m_1)g = (m_2 + m_1)a\)

Solving for acceleration \(a\):

\(a = \frac{(m_2 - m_1)g}{m_2 + m_1}\)

If the problem states or implies that \(m_2 = 2m_1\) (a common scenario for illustrative purposes), the calculation is simplified:

\(a = \frac{(2m_1 - m_1)g}{2m_1 + m_1} = \frac{m_1g}{3m_1} = \frac{g}{3}\)

Therefore, the correct answer is:

\(\frac{g}{3}\)