JEE MAIN - Physics (2022 - 30th June Morning Shift - No. 6)
A 2 kg block is pushed against a vertical wall by applying a horizontal force of 50 N. The coefficient of static friction between the block and the wall is 0.5. A force F is also applied on the block vertically upward (as shown in figure). The maximum value of F applied, so that the block does not move upward, will be :
(Given : g = 10 ms$$-$$2)
_30th_June_Morning_Shift_en_6_1.png)
10 N
20 N
25 N
45 N
Explanation
_30th_June_Morning_Shift_en_6_2.png)
Here Fmax force is trying to move the block upward so friction force will be applied towards downward.
Along horizontal direction,
N = 50 N
Along vertical diretion,
Fmax = mg + fmax
= 2 $$\times$$ 10 + $$\mu$$N
= 20 + 0.5 $$\times$$ 50
= 20 + 25
= 45 N
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