JEE Main 2019 — Physics solution

JEE Main2019Physics

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 and between B and the surface of the table is also 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: [Take g = 10 m/s²]

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

Solution

$a_{c} = (\frac{F - f}{M + m}) a = \frac{F - ( 0.2 ) 4 \times 10}{4} = (\frac{F - 8}{4}) W e ha v e \frac{F - 8}{4} \leq (0.2) 10 \Rightarrow F - 8 \leq 8 \Rightarrow F \leq 16$

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