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 the table and from its other end, another block B of mass $m_2$ is suspended. The coefficient of kinetic friction between block A and the table is $\mu_k$. When block A is sliding on the table, the tension in the string is:

TopperSquare Expert · Accepted Answer

The correct answer is Option (D): $\frac{m_1 m_2 (1 + \mu_k)g}{(m_1 + m_2)}$. 1. **System Analysis:** Block B ($m_2$) moves vertically downwards and block A ($m_1$) moves horizontally. Both share the same magnitude of acceleration $a$ because they are connected by an inextensible string.
2. **Equations of Motion:**
   - **For Block B ($m_2$):** The forces are weight $m_2g$ (down) and tension $T$ (up). Since it accelerates down:
     $$m_2g - T = m_2a \quad 	ext{...(i)}$$
   - **For Block A ($m_1$):** The forces are tension $T$ (forward) and kinetic friction $f_k$ (backward). The vertical forces balance ($N = m_1g$), so $f_k = \mu_k N = \mu_k m_1g$. Since it accelerates forward:
     $$T - f_k = m_1a \Rightarrow T - \mu_k m_1g = m_1a \quad 	ext{...(ii)}$$
3. **Finding Acceleration:** Add equations (i) and (ii) to eliminate $T$:
   $$m_2g - \mu_k m_1g = (m_1 + m_2)a$$
   $$a = \frac{g(m_2 - \mu_k m_1)}{m_1 + m_2}$$
4. **Finding Tension:** Substitute $a$ back into equation (ii):
   $$T = m_1a + \mu_k m_1g$$
   $$T = m_1 \left[ \frac{g(m_2 - \mu_k m_1)}{m_1 + m_2} ight] + \mu_k m_1g$$
   $$T = \frac{m_1 m_2 g - \mu_k m_1^2 g + \mu_k m_1^2 g + \mu_k m_1 m_2 g}{m_1 + m_2}$$
   $$T = \frac{m_1 m_2 g (1 + \mu_k)}{m_1 + m_2}$$
   (Reference: NCERT Class 11, Physics Part I, Chapter 5, Section 5.10 and Example 5.9).

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