JEE MAIN - Physics (2022 - 27th June Morning Shift - No. 4)
A system of two blocks of masses m = 2 kg and M = 8 kg is placed on a smooth table as shown in figure. The coefficient of static friction between two blocks is 0.5. The maximum horizontal force F that can be applied to the block of mass M so that the blocks move together will be :
_27th_June_Morning_Shift_en_4_1.png)
9.8 N
39.2 N
49 N
78.4 N
Explanation
_27th_June_Morning_Shift_en_4_2.png)
$$\therefore$$ amax = $$\mu$$g
= 0.5 $$\times$$ 9.8 = 4.9 m/s2
$$\therefore$$ Fmax = (8 + 2) $$\times$$ 4.9 = 49 N
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