Question

A block A of mass $m_1$ rests on a horizontal table. A light string connected to it passes over a frictionless pully at the edge of table and from its other end another block B of mass $m_2$ is suspended. The coefficient of kinetic friction between the block and the table is $\mu_k$. When the block A is sliding on the table, the tension in the string is

• A. $\frac{m_1m_2(1+\mu_k)g}{(m_1+m_2)}$
• B. $\frac{m_1m_2(1-\mu_k)g}{(m_1+m_2)}$
• C. $\frac{(m_1+\mu_k m_1)g}{(m_1+m_2)}$
• D. $\frac{(m_2-\mu_k m_1)g}{(m_1+m_2)}$

Answer 1

The Correct Option is A

To determine the tension in the string when block A is sliding on the table, we begin by analyzing the forces acting on both blocks.

1. Identify Forces on Block A:
   • Weight of Block A: m_1g
   • Normal Force: N = m_1g (as it is on a horizontal table)
   • Frictional Force: f_k = \mu_k m_1 g
   • Tension in the string: T
2. Apply Newton's Second Law to Block A:

   The net force on Block A can be written as:

   m_1 a = T - \mu_k m_1 g (Equation 1)

3. Identify Forces on Block B:
   • Weight of Block B: m_2g
   • Tension in the string: T
4. Apply Newton's Second Law to Block B:

   The net force on Block B is due to gravity minus the tension:

   m_2 a = m_2 g - T (Equation 2)

5. Solve the Two Equations Simultaneously:
   • From Equation 1: T = m_1 a + \mu_k m_1 g
   • From Equation 2: m_2 g - T = m_2 a \Rightarrow T = m_2 g - m_2 a

   Equate the two expressions for T:

   m_1 a + \mu_k m_1 g = m_2 g - m_2 a

   Rearrange to solve for a:

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

   a = \frac{m_2 g - \mu_k m_1 g}{m_1 + m_2}

6. Substitute a back to Find T:

   T = m_1 a + \mu_k m_1 g

   Substitute a:

   T = m_1 \left(\frac{m_2 g - \mu_k m_1 g}{m_1 + m_2}\right) + \mu_k m_1 g

   Simplify to find:

   T = \frac{m_1 m_2 g}{m_1 + m_2} - \frac{\mu_k m_1^2 g}{m_1 + m_2} + \frac{\mu_k m_1 m_1 g}{m_1 + m_2}

   Combine terms:

   T = \frac{m_1 m_2 g + \mu_k m_1 m_2 g}{m_1 + m_2}

   So, the tension in the string is:

   T = \frac{m_1 m_2 (1 + \mu_k) g}{m_1 + m_2}

Therefore, the correct answer is \frac{m_1m_2(1+\mu_k)g}{(m_1+m_2)}, which is option A.