A block A of mass $m_1$ rests on a horizontal table. A light string connected to it passes over a frictionless

Question

A block A of mass $m_1$ rests on a horizontal table. A light string connected to it passes over a frictionless pulley 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:

Accepted Answer

1. **Free Body Diagram of Block B ($m_2$):** 
   - Forces acting are gravity ($m_2g$) downwards and tension ($T$) upwards.
   - Equation of motion: $m_2g - T = m_2a$ ... (i) (assuming acceleration $a$ downwards).
2. **Free Body Diagram of Block A ($m_1$):** 
   - Forces acting horizontally are tension ($T$) towards the pulley and kinetic friction ($f_k$) opposing motion.
   - Vertical forces balance: $N = m_1g$, so $f_k = \mu_k N = \mu_k m_1g$.
   - Equation of motion: $T - f_k = m_1a \implies T - \mu_k m_1g = m_1a$ ... (ii).
3. **Solve for Acceleration ($a$):** 
   - Adding (i) and (ii): $m_2g - \mu_k m_1g = (m_1 + m_2)a$.
   - $a = \frac{g(m_2 - \mu_k m_1)}{m_1 + m_2}$.
4. **Solve for Tension ($T$):** 
   - Substitute $a$ into equation (ii): $T = m_1(a + \mu_k g)$.
   - $T = m_1 \left[ \frac{g(m_2 - \mu_k m_1)}{m_1 + m_2} + \mu_k g \right]$.
   - $T = \frac{m_1 g}{m_1 + m_2} [ m_2 - \mu_k m_1 + \mu_k(m_1 + m_2) ]$.
   - $T = \frac{m_1 g}{m_1 + m_2} [ m_2 - \mu_k m_1 + \mu_k m_1 + \mu_k m_2 ]$.
   - $T = \frac{m_1 g}{m_1 + m_2} [ m_2 + \mu_k m_2 ] = \frac{m_1 m_2 g (1 + \mu_k)}{m_1 + m_2}$.
   (Reference: NCERT Class 11, Physics Part I, Chapter 5: Laws of Motion, Section 5.9 Friction).

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