Verified Solution • 2025–26 Session
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QJEEM19Ph80b9: JEE Main 2019 Physics - Class 11 JEEM19Ph80b9 Solution: Two blocks A and B of masses m A = 1 kg and m B = 3 kg are kept …
Two blocks A and B of masses mA = 1 kg and mB = 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]
[Take g = 10 m/s2]
Multiple Choice Options
- A 8 N
- B 16 N
- C 40 N
- D 12 N
Step-by-Step Verified Solution
MA = 1 kg, MB = 3 k
Fmax = (MA + MB) × 0.2 × 10 + (MA + MB) × 0.2 × 10
= 4 × 2 + 4 × 2 = 16
Fmax = (MA + MB) × 0.2 × 10 + (MA + MB) × 0.2 × 10
= 4 × 2 + 4 × 2 = 16