JEE Main 2019 — Laws Of Motion Question with Solution
From: JEE Main 2019 (Online) 10th April Evening Slot
Question
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]
This question includes a diagram. The text above accompanies the figure.
Choose an option
Show full solutionCorrect option: B
Correct answer
B16 N
Step-by-step explanation
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
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