Question

Two blocks $A$ and $B$ of masses $m_A = 1$ kg and $m_B = 3$ kg are kept on the table as shown in figure. The coefficient of friction between $A$ and $B$ is $0.2$. The maximum force $F$ that can be applied on $B$ horizontally, so that the block $A$ does not slide over the block $B$ is (2019) [Take $g = 10 m/s^2$]

• A. 8 N
• B. 16 N
• C. 40 N
• D. 12 N

Answer 1

Answer: (A)

8 N

Answer 2

1. For block A not to slip on block B, its acceleration must not exceed

   $a_{\text{max}} = \mu g = 0.2 \times 10 = 2 \, \text{m/s}^2.$

2. To accelerate block A at $2 \, \text{m/s}^2$, the friction force required is

   $f = m_A a = 1 \times 2 = 2 \, \text{N}.$

3. The entire system (blocks A and B) accelerates together. For block B, applying Newton’s second law yields

   $F - f = m_B \, a.$

4. Therefore,

   $F = m_B a + f = 3 \times 2 + 2 = 6 + 2 = 8 \, \text{N}.$

Thus, the maximum force $F$ that can be applied on block B so that block A does not slide is 8 N.