JEE MAIN - Physics (2019 - 10th April Evening Slot - No. 5)
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]_10th_April_Evening_Slot_en_5_1.png)
[Take g = 10 m/s2]
8 N
16 N
40 N
12 N
Explanation
MA = 1 kg, MB = 3 k
$${\mu _{AB}} = 0.2$$
$${\mu _B} = 0.2$$
Fmax = (MA + MB) × 0.2 × 10 + (MA + MB) × 0.2 × 10
= 4 × 2 + 4 × 2 = 16
$${\mu _{AB}} = 0.2$$
$${\mu _B} = 0.2$$
Fmax = (MA + MB) × 0.2 × 10 + (MA + MB) × 0.2 × 10
= 4 × 2 + 4 × 2 = 16
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