JEE PYQ: Laws Of Motion Question 25
Question 25 - 2019 | 10 Apr Shift 2] (MCQ)
Two blocks A and B masses $m_A = 1$ kg and $m_B = 3$ kg are kept on the table. The coefficient of friction between A and B is 0.2 and between B and the surface is also 0.2. The maximum force F that can be applied on B horizontally, so that block A does not slide over B is :
(1) 8 N
(2) 16 N [Correct]
(3) 40 N
(4) 12 N
Show Answer
Answer: (2)
Solution
$a_{\max} = \mu g = 2,\text{m/s}^2$. $F - 8 = 4 \times 2$. $F = 16$ N.
BETA
